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Why can I use a list index as an indexing variable in a for loop?
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I have the following code:
a = [0,1,2,3]
for a[-1] in a:
print(a[-1])
The output is:
0,1,2,2
I'm confused about why a list index can be used as an indexing variable in a for a loop?
python
New contributor
|
show 1 more comment
I have the following code:
a = [0,1,2,3]
for a[-1] in a:
print(a[-1])
The output is:
0,1,2,2
I'm confused about why a list index can be used as an indexing variable in a for a loop?
python
New contributor
6
Somehow this question looks like a bad and newbie question, but I don't get the logic either lol
– gameon67
38 mins ago
3
I don't know why you would ever want to do this, but now I know you can
– Nathan
35 mins ago
over the iteration only the last time a[-1] gets its value, all other times python treats as iteration variable.
– Arun Augustine
32 mins ago
6
This would be a great question for an awful coding interview
– Nathan
29 mins ago
@Nathan same idea, would be a great trivial question in interview
– gameon67
28 mins ago
|
show 1 more comment
I have the following code:
a = [0,1,2,3]
for a[-1] in a:
print(a[-1])
The output is:
0,1,2,2
I'm confused about why a list index can be used as an indexing variable in a for a loop?
python
New contributor
I have the following code:
a = [0,1,2,3]
for a[-1] in a:
print(a[-1])
The output is:
0,1,2,2
I'm confused about why a list index can be used as an indexing variable in a for a loop?
python
python
New contributor
New contributor
edited 7 mins ago
Christian Dean
15.8k62859
15.8k62859
New contributor
asked 47 mins ago
Kundan VermaKundan Verma
611
611
New contributor
New contributor
6
Somehow this question looks like a bad and newbie question, but I don't get the logic either lol
– gameon67
38 mins ago
3
I don't know why you would ever want to do this, but now I know you can
– Nathan
35 mins ago
over the iteration only the last time a[-1] gets its value, all other times python treats as iteration variable.
– Arun Augustine
32 mins ago
6
This would be a great question for an awful coding interview
– Nathan
29 mins ago
@Nathan same idea, would be a great trivial question in interview
– gameon67
28 mins ago
|
show 1 more comment
6
Somehow this question looks like a bad and newbie question, but I don't get the logic either lol
– gameon67
38 mins ago
3
I don't know why you would ever want to do this, but now I know you can
– Nathan
35 mins ago
over the iteration only the last time a[-1] gets its value, all other times python treats as iteration variable.
– Arun Augustine
32 mins ago
6
This would be a great question for an awful coding interview
– Nathan
29 mins ago
@Nathan same idea, would be a great trivial question in interview
– gameon67
28 mins ago
6
6
Somehow this question looks like a bad and newbie question, but I don't get the logic either lol
– gameon67
38 mins ago
Somehow this question looks like a bad and newbie question, but I don't get the logic either lol
– gameon67
38 mins ago
3
3
I don't know why you would ever want to do this, but now I know you can
– Nathan
35 mins ago
I don't know why you would ever want to do this, but now I know you can
– Nathan
35 mins ago
over the iteration only the last time a[-1] gets its value, all other times python treats as iteration variable.
– Arun Augustine
32 mins ago
over the iteration only the last time a[-1] gets its value, all other times python treats as iteration variable.
– Arun Augustine
32 mins ago
6
6
This would be a great question for an awful coding interview
– Nathan
29 mins ago
This would be a great question for an awful coding interview
– Nathan
29 mins ago
@Nathan same idea, would be a great trivial question in interview
– gameon67
28 mins ago
@Nathan same idea, would be a great trivial question in interview
– gameon67
28 mins ago
|
show 1 more comment
5 Answers
5
active
oldest
votes
To bring a language-lawyer aspect to the question, let's look at documentation:
for_stmt ::= "for" target_list "in" expression_list ":" suite
The expression list is evaluated once; it should yield an iterable object. An iterator is created for the result of the
expression_list
. The suite is then executed once for each item provided by the iterator, in the order returned by the iterator. Each item in turn is assigned to the target list using the standard rules for assignments (see Assignment statements), and then the suite is executed.
(emphasis not originally in docs)
The suite refers to the statements under the for-block. print(a[-1])
in OP's example.
Simply, on each iteration, the loop variable (target_list
) gets assigned to the next item in the iterable (expression_list
).
Let's extend the print statement:
a = [0, 1, 2, 3]
for a[-1] in a:
print(a, a[-1])
This gives the following output:
[0, 1, 2, 0] 0 # a[-1] assigned 0
[0, 1, 2, 1] 1 # a[-1] assigned 1
[0, 1, 2, 2] 2 # a[-1] assigned 2
[0, 1, 2, 2] 2 # a[-1] assigned itself (2)
As a[-1]
is a valid form left-hand side, assignments to a[-1]
will mutate a
, modifying the list during iteration. In this particular example, a[-1]
retains -2
from the previous evaluation. If we trace the assignment on each iteration, we have
a[-1] = 0 # a[0] # a[3] is 0 now
a[-1] = 1 # a[1] # a[3] is 1 now
a[-1] = 2 # a[2] # a[3] is 2 now
a[-1] = 2 # a[3] # a[3] is itself (2)
2
Might be useful to bold Each item [...] is assigned to the target list using the standard rules for assignments.
– Mateen Ulhaq
23 mins ago
2
Essentially, the important thing to note here is that Python considersa[-1]
to be a valid form of the left-hand side of an assignment statement (e.g.a[-1] = 1
is valid grammar). Thusa[-1]
is a valid "variable" name, because as the documentation stated, it evaluates the binding variable(s) in a for loop declaration as it would the left-hand side of an assignment.
– Christian Dean
12 mins ago
add a comment |
it is an interesting question, you can understand it by that:
for v in a:
a[-1] = v
print(a[-1])
print(a)
actually a
becomes: [0, 1, 2, 2]
after loop
output:
0
1
2
2
[0, 1, 2, 2]
Hope that helps you, and comment if you have further questions. : )
add a comment |
(this is more of a long comment than an answer - there are a couple of good ones already, especially @TrebledJ's. But I had to think of it explicitly in terms of overwriting variables that already have values before it clicked for me.)
If you had
x = 0
l = [1, 2, 3]
for x in l:
print(x)
you wouldn't be surprised that x
is overriden each time through the loop. Even though x
existed before, its value isn't used (i.e. for 0 in l:
, which would throw an error). Rather, we assign the values from l
to x
.
When we do
a = [0, 1, 2, 3]
for a[-1] in a:
print(a[-1])
even though a[-1]
already exists and has a value, we don't put that value in but rather assign to a[-1]
each time through the loop.
Somehow I thought the variable used infor loop
is immutable. +1 for pointing it out.
– Amir A. Shabani
15 mins ago
add a comment |
a[-1]
refers to the last element of a
, in this case a[3]
. The for
loop is a bit unusual in that it is using this element as the loop variable. So it first gets set to 0, then 1, then 2. Finally, on the last iteration, the for
loop retrieves a[3]
which at that point is 2
, so the list ends up as [0, 1, 2, 2]
.
A more typical for
loop uses a simple local variable name as the loop variable, e.g. for x ...
. In that case, x
is set to the next value upon each iteration. This case is no different, except that a[-1]
is set to the next value upon each iteration. You don't see this very often, but it's consistent.
How does it actuallygets set to 0, then 1, then 2 and finally 2
? That's the confusing part!
– Amir A. Shabani
40 mins ago
1
Agree, I don't really understand by reading this answer
– gameon67
39 mins ago
I expected it would be like sayingfor 3 in a: print(a[-1])
(becausea[-1]
was3
at the start of the loop) and give an error, but clearly that's not the case. You're obviously correct about what's happening, but I'm surprised this evaluates this way.
– Nathan
36 mins ago
It's set by thefor
loop. Each time through, it gets set to the next element of the list. What's the first element? Right, it's 0. What's the second? Right again, it's 1. etc.
– Tom Karzes
36 mins ago
1
"The for loop is a bit unusual in that it is using this element as the loop variable." - This may be where the confusion is coming from. Python isn't using the elementa[-1]
evaluates to, rather, it's using the indexa[-1]
itself. If you rephrase that statement, your answer becomes much more clear.
– Christian Dean
3 mins ago
|
show 1 more comment
The left expression of a for
loop statement gets assigned with each item in the iterable on the right in each iteration, so
for n in a:
print(n)
is just a fancy way of doing:
for i in range(len(a)):
n = a[i]
print(n)
Likewise,
for a[-1] in a:
print(a[-1])
is just a fancy way of doing:
for i in range(len(a)):
a[-1] = a[i]
print(a[-1])
where in each iteration, the last item of a
gets assigned with the next item in a
, so when the iteration finally comes to the last item, its value got last assigned with the second-last item, 2
.
add a comment |
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5 Answers
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5 Answers
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oldest
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To bring a language-lawyer aspect to the question, let's look at documentation:
for_stmt ::= "for" target_list "in" expression_list ":" suite
The expression list is evaluated once; it should yield an iterable object. An iterator is created for the result of the
expression_list
. The suite is then executed once for each item provided by the iterator, in the order returned by the iterator. Each item in turn is assigned to the target list using the standard rules for assignments (see Assignment statements), and then the suite is executed.
(emphasis not originally in docs)
The suite refers to the statements under the for-block. print(a[-1])
in OP's example.
Simply, on each iteration, the loop variable (target_list
) gets assigned to the next item in the iterable (expression_list
).
Let's extend the print statement:
a = [0, 1, 2, 3]
for a[-1] in a:
print(a, a[-1])
This gives the following output:
[0, 1, 2, 0] 0 # a[-1] assigned 0
[0, 1, 2, 1] 1 # a[-1] assigned 1
[0, 1, 2, 2] 2 # a[-1] assigned 2
[0, 1, 2, 2] 2 # a[-1] assigned itself (2)
As a[-1]
is a valid form left-hand side, assignments to a[-1]
will mutate a
, modifying the list during iteration. In this particular example, a[-1]
retains -2
from the previous evaluation. If we trace the assignment on each iteration, we have
a[-1] = 0 # a[0] # a[3] is 0 now
a[-1] = 1 # a[1] # a[3] is 1 now
a[-1] = 2 # a[2] # a[3] is 2 now
a[-1] = 2 # a[3] # a[3] is itself (2)
2
Might be useful to bold Each item [...] is assigned to the target list using the standard rules for assignments.
– Mateen Ulhaq
23 mins ago
2
Essentially, the important thing to note here is that Python considersa[-1]
to be a valid form of the left-hand side of an assignment statement (e.g.a[-1] = 1
is valid grammar). Thusa[-1]
is a valid "variable" name, because as the documentation stated, it evaluates the binding variable(s) in a for loop declaration as it would the left-hand side of an assignment.
– Christian Dean
12 mins ago
add a comment |
To bring a language-lawyer aspect to the question, let's look at documentation:
for_stmt ::= "for" target_list "in" expression_list ":" suite
The expression list is evaluated once; it should yield an iterable object. An iterator is created for the result of the
expression_list
. The suite is then executed once for each item provided by the iterator, in the order returned by the iterator. Each item in turn is assigned to the target list using the standard rules for assignments (see Assignment statements), and then the suite is executed.
(emphasis not originally in docs)
The suite refers to the statements under the for-block. print(a[-1])
in OP's example.
Simply, on each iteration, the loop variable (target_list
) gets assigned to the next item in the iterable (expression_list
).
Let's extend the print statement:
a = [0, 1, 2, 3]
for a[-1] in a:
print(a, a[-1])
This gives the following output:
[0, 1, 2, 0] 0 # a[-1] assigned 0
[0, 1, 2, 1] 1 # a[-1] assigned 1
[0, 1, 2, 2] 2 # a[-1] assigned 2
[0, 1, 2, 2] 2 # a[-1] assigned itself (2)
As a[-1]
is a valid form left-hand side, assignments to a[-1]
will mutate a
, modifying the list during iteration. In this particular example, a[-1]
retains -2
from the previous evaluation. If we trace the assignment on each iteration, we have
a[-1] = 0 # a[0] # a[3] is 0 now
a[-1] = 1 # a[1] # a[3] is 1 now
a[-1] = 2 # a[2] # a[3] is 2 now
a[-1] = 2 # a[3] # a[3] is itself (2)
2
Might be useful to bold Each item [...] is assigned to the target list using the standard rules for assignments.
– Mateen Ulhaq
23 mins ago
2
Essentially, the important thing to note here is that Python considersa[-1]
to be a valid form of the left-hand side of an assignment statement (e.g.a[-1] = 1
is valid grammar). Thusa[-1]
is a valid "variable" name, because as the documentation stated, it evaluates the binding variable(s) in a for loop declaration as it would the left-hand side of an assignment.
– Christian Dean
12 mins ago
add a comment |
To bring a language-lawyer aspect to the question, let's look at documentation:
for_stmt ::= "for" target_list "in" expression_list ":" suite
The expression list is evaluated once; it should yield an iterable object. An iterator is created for the result of the
expression_list
. The suite is then executed once for each item provided by the iterator, in the order returned by the iterator. Each item in turn is assigned to the target list using the standard rules for assignments (see Assignment statements), and then the suite is executed.
(emphasis not originally in docs)
The suite refers to the statements under the for-block. print(a[-1])
in OP's example.
Simply, on each iteration, the loop variable (target_list
) gets assigned to the next item in the iterable (expression_list
).
Let's extend the print statement:
a = [0, 1, 2, 3]
for a[-1] in a:
print(a, a[-1])
This gives the following output:
[0, 1, 2, 0] 0 # a[-1] assigned 0
[0, 1, 2, 1] 1 # a[-1] assigned 1
[0, 1, 2, 2] 2 # a[-1] assigned 2
[0, 1, 2, 2] 2 # a[-1] assigned itself (2)
As a[-1]
is a valid form left-hand side, assignments to a[-1]
will mutate a
, modifying the list during iteration. In this particular example, a[-1]
retains -2
from the previous evaluation. If we trace the assignment on each iteration, we have
a[-1] = 0 # a[0] # a[3] is 0 now
a[-1] = 1 # a[1] # a[3] is 1 now
a[-1] = 2 # a[2] # a[3] is 2 now
a[-1] = 2 # a[3] # a[3] is itself (2)
To bring a language-lawyer aspect to the question, let's look at documentation:
for_stmt ::= "for" target_list "in" expression_list ":" suite
The expression list is evaluated once; it should yield an iterable object. An iterator is created for the result of the
expression_list
. The suite is then executed once for each item provided by the iterator, in the order returned by the iterator. Each item in turn is assigned to the target list using the standard rules for assignments (see Assignment statements), and then the suite is executed.
(emphasis not originally in docs)
The suite refers to the statements under the for-block. print(a[-1])
in OP's example.
Simply, on each iteration, the loop variable (target_list
) gets assigned to the next item in the iterable (expression_list
).
Let's extend the print statement:
a = [0, 1, 2, 3]
for a[-1] in a:
print(a, a[-1])
This gives the following output:
[0, 1, 2, 0] 0 # a[-1] assigned 0
[0, 1, 2, 1] 1 # a[-1] assigned 1
[0, 1, 2, 2] 2 # a[-1] assigned 2
[0, 1, 2, 2] 2 # a[-1] assigned itself (2)
As a[-1]
is a valid form left-hand side, assignments to a[-1]
will mutate a
, modifying the list during iteration. In this particular example, a[-1]
retains -2
from the previous evaluation. If we trace the assignment on each iteration, we have
a[-1] = 0 # a[0] # a[3] is 0 now
a[-1] = 1 # a[1] # a[3] is 1 now
a[-1] = 2 # a[2] # a[3] is 2 now
a[-1] = 2 # a[3] # a[3] is itself (2)
edited 8 mins ago
answered 27 mins ago
TrebledJTrebledJ
3,74421328
3,74421328
2
Might be useful to bold Each item [...] is assigned to the target list using the standard rules for assignments.
– Mateen Ulhaq
23 mins ago
2
Essentially, the important thing to note here is that Python considersa[-1]
to be a valid form of the left-hand side of an assignment statement (e.g.a[-1] = 1
is valid grammar). Thusa[-1]
is a valid "variable" name, because as the documentation stated, it evaluates the binding variable(s) in a for loop declaration as it would the left-hand side of an assignment.
– Christian Dean
12 mins ago
add a comment |
2
Might be useful to bold Each item [...] is assigned to the target list using the standard rules for assignments.
– Mateen Ulhaq
23 mins ago
2
Essentially, the important thing to note here is that Python considersa[-1]
to be a valid form of the left-hand side of an assignment statement (e.g.a[-1] = 1
is valid grammar). Thusa[-1]
is a valid "variable" name, because as the documentation stated, it evaluates the binding variable(s) in a for loop declaration as it would the left-hand side of an assignment.
– Christian Dean
12 mins ago
2
2
Might be useful to bold Each item [...] is assigned to the target list using the standard rules for assignments.
– Mateen Ulhaq
23 mins ago
Might be useful to bold Each item [...] is assigned to the target list using the standard rules for assignments.
– Mateen Ulhaq
23 mins ago
2
2
Essentially, the important thing to note here is that Python considers
a[-1]
to be a valid form of the left-hand side of an assignment statement (e.g. a[-1] = 1
is valid grammar). Thus a[-1]
is a valid "variable" name, because as the documentation stated, it evaluates the binding variable(s) in a for loop declaration as it would the left-hand side of an assignment.– Christian Dean
12 mins ago
Essentially, the important thing to note here is that Python considers
a[-1]
to be a valid form of the left-hand side of an assignment statement (e.g. a[-1] = 1
is valid grammar). Thus a[-1]
is a valid "variable" name, because as the documentation stated, it evaluates the binding variable(s) in a for loop declaration as it would the left-hand side of an assignment.– Christian Dean
12 mins ago
add a comment |
it is an interesting question, you can understand it by that:
for v in a:
a[-1] = v
print(a[-1])
print(a)
actually a
becomes: [0, 1, 2, 2]
after loop
output:
0
1
2
2
[0, 1, 2, 2]
Hope that helps you, and comment if you have further questions. : )
add a comment |
it is an interesting question, you can understand it by that:
for v in a:
a[-1] = v
print(a[-1])
print(a)
actually a
becomes: [0, 1, 2, 2]
after loop
output:
0
1
2
2
[0, 1, 2, 2]
Hope that helps you, and comment if you have further questions. : )
add a comment |
it is an interesting question, you can understand it by that:
for v in a:
a[-1] = v
print(a[-1])
print(a)
actually a
becomes: [0, 1, 2, 2]
after loop
output:
0
1
2
2
[0, 1, 2, 2]
Hope that helps you, and comment if you have further questions. : )
it is an interesting question, you can understand it by that:
for v in a:
a[-1] = v
print(a[-1])
print(a)
actually a
becomes: [0, 1, 2, 2]
after loop
output:
0
1
2
2
[0, 1, 2, 2]
Hope that helps you, and comment if you have further questions. : )
answered 29 mins ago
recnacrecnac
1,183123
1,183123
add a comment |
add a comment |
(this is more of a long comment than an answer - there are a couple of good ones already, especially @TrebledJ's. But I had to think of it explicitly in terms of overwriting variables that already have values before it clicked for me.)
If you had
x = 0
l = [1, 2, 3]
for x in l:
print(x)
you wouldn't be surprised that x
is overriden each time through the loop. Even though x
existed before, its value isn't used (i.e. for 0 in l:
, which would throw an error). Rather, we assign the values from l
to x
.
When we do
a = [0, 1, 2, 3]
for a[-1] in a:
print(a[-1])
even though a[-1]
already exists and has a value, we don't put that value in but rather assign to a[-1]
each time through the loop.
Somehow I thought the variable used infor loop
is immutable. +1 for pointing it out.
– Amir A. Shabani
15 mins ago
add a comment |
(this is more of a long comment than an answer - there are a couple of good ones already, especially @TrebledJ's. But I had to think of it explicitly in terms of overwriting variables that already have values before it clicked for me.)
If you had
x = 0
l = [1, 2, 3]
for x in l:
print(x)
you wouldn't be surprised that x
is overriden each time through the loop. Even though x
existed before, its value isn't used (i.e. for 0 in l:
, which would throw an error). Rather, we assign the values from l
to x
.
When we do
a = [0, 1, 2, 3]
for a[-1] in a:
print(a[-1])
even though a[-1]
already exists and has a value, we don't put that value in but rather assign to a[-1]
each time through the loop.
Somehow I thought the variable used infor loop
is immutable. +1 for pointing it out.
– Amir A. Shabani
15 mins ago
add a comment |
(this is more of a long comment than an answer - there are a couple of good ones already, especially @TrebledJ's. But I had to think of it explicitly in terms of overwriting variables that already have values before it clicked for me.)
If you had
x = 0
l = [1, 2, 3]
for x in l:
print(x)
you wouldn't be surprised that x
is overriden each time through the loop. Even though x
existed before, its value isn't used (i.e. for 0 in l:
, which would throw an error). Rather, we assign the values from l
to x
.
When we do
a = [0, 1, 2, 3]
for a[-1] in a:
print(a[-1])
even though a[-1]
already exists and has a value, we don't put that value in but rather assign to a[-1]
each time through the loop.
(this is more of a long comment than an answer - there are a couple of good ones already, especially @TrebledJ's. But I had to think of it explicitly in terms of overwriting variables that already have values before it clicked for me.)
If you had
x = 0
l = [1, 2, 3]
for x in l:
print(x)
you wouldn't be surprised that x
is overriden each time through the loop. Even though x
existed before, its value isn't used (i.e. for 0 in l:
, which would throw an error). Rather, we assign the values from l
to x
.
When we do
a = [0, 1, 2, 3]
for a[-1] in a:
print(a[-1])
even though a[-1]
already exists and has a value, we don't put that value in but rather assign to a[-1]
each time through the loop.
edited 12 mins ago
Amir A. Shabani
455616
455616
answered 19 mins ago
NathanNathan
2,02511326
2,02511326
Somehow I thought the variable used infor loop
is immutable. +1 for pointing it out.
– Amir A. Shabani
15 mins ago
add a comment |
Somehow I thought the variable used infor loop
is immutable. +1 for pointing it out.
– Amir A. Shabani
15 mins ago
Somehow I thought the variable used in
for loop
is immutable. +1 for pointing it out.– Amir A. Shabani
15 mins ago
Somehow I thought the variable used in
for loop
is immutable. +1 for pointing it out.– Amir A. Shabani
15 mins ago
add a comment |
a[-1]
refers to the last element of a
, in this case a[3]
. The for
loop is a bit unusual in that it is using this element as the loop variable. So it first gets set to 0, then 1, then 2. Finally, on the last iteration, the for
loop retrieves a[3]
which at that point is 2
, so the list ends up as [0, 1, 2, 2]
.
A more typical for
loop uses a simple local variable name as the loop variable, e.g. for x ...
. In that case, x
is set to the next value upon each iteration. This case is no different, except that a[-1]
is set to the next value upon each iteration. You don't see this very often, but it's consistent.
How does it actuallygets set to 0, then 1, then 2 and finally 2
? That's the confusing part!
– Amir A. Shabani
40 mins ago
1
Agree, I don't really understand by reading this answer
– gameon67
39 mins ago
I expected it would be like sayingfor 3 in a: print(a[-1])
(becausea[-1]
was3
at the start of the loop) and give an error, but clearly that's not the case. You're obviously correct about what's happening, but I'm surprised this evaluates this way.
– Nathan
36 mins ago
It's set by thefor
loop. Each time through, it gets set to the next element of the list. What's the first element? Right, it's 0. What's the second? Right again, it's 1. etc.
– Tom Karzes
36 mins ago
1
"The for loop is a bit unusual in that it is using this element as the loop variable." - This may be where the confusion is coming from. Python isn't using the elementa[-1]
evaluates to, rather, it's using the indexa[-1]
itself. If you rephrase that statement, your answer becomes much more clear.
– Christian Dean
3 mins ago
|
show 1 more comment
a[-1]
refers to the last element of a
, in this case a[3]
. The for
loop is a bit unusual in that it is using this element as the loop variable. So it first gets set to 0, then 1, then 2. Finally, on the last iteration, the for
loop retrieves a[3]
which at that point is 2
, so the list ends up as [0, 1, 2, 2]
.
A more typical for
loop uses a simple local variable name as the loop variable, e.g. for x ...
. In that case, x
is set to the next value upon each iteration. This case is no different, except that a[-1]
is set to the next value upon each iteration. You don't see this very often, but it's consistent.
How does it actuallygets set to 0, then 1, then 2 and finally 2
? That's the confusing part!
– Amir A. Shabani
40 mins ago
1
Agree, I don't really understand by reading this answer
– gameon67
39 mins ago
I expected it would be like sayingfor 3 in a: print(a[-1])
(becausea[-1]
was3
at the start of the loop) and give an error, but clearly that's not the case. You're obviously correct about what's happening, but I'm surprised this evaluates this way.
– Nathan
36 mins ago
It's set by thefor
loop. Each time through, it gets set to the next element of the list. What's the first element? Right, it's 0. What's the second? Right again, it's 1. etc.
– Tom Karzes
36 mins ago
1
"The for loop is a bit unusual in that it is using this element as the loop variable." - This may be where the confusion is coming from. Python isn't using the elementa[-1]
evaluates to, rather, it's using the indexa[-1]
itself. If you rephrase that statement, your answer becomes much more clear.
– Christian Dean
3 mins ago
|
show 1 more comment
a[-1]
refers to the last element of a
, in this case a[3]
. The for
loop is a bit unusual in that it is using this element as the loop variable. So it first gets set to 0, then 1, then 2. Finally, on the last iteration, the for
loop retrieves a[3]
which at that point is 2
, so the list ends up as [0, 1, 2, 2]
.
A more typical for
loop uses a simple local variable name as the loop variable, e.g. for x ...
. In that case, x
is set to the next value upon each iteration. This case is no different, except that a[-1]
is set to the next value upon each iteration. You don't see this very often, but it's consistent.
a[-1]
refers to the last element of a
, in this case a[3]
. The for
loop is a bit unusual in that it is using this element as the loop variable. So it first gets set to 0, then 1, then 2. Finally, on the last iteration, the for
loop retrieves a[3]
which at that point is 2
, so the list ends up as [0, 1, 2, 2]
.
A more typical for
loop uses a simple local variable name as the loop variable, e.g. for x ...
. In that case, x
is set to the next value upon each iteration. This case is no different, except that a[-1]
is set to the next value upon each iteration. You don't see this very often, but it's consistent.
edited 28 mins ago
answered 42 mins ago
Tom KarzesTom Karzes
11.2k1926
11.2k1926
How does it actuallygets set to 0, then 1, then 2 and finally 2
? That's the confusing part!
– Amir A. Shabani
40 mins ago
1
Agree, I don't really understand by reading this answer
– gameon67
39 mins ago
I expected it would be like sayingfor 3 in a: print(a[-1])
(becausea[-1]
was3
at the start of the loop) and give an error, but clearly that's not the case. You're obviously correct about what's happening, but I'm surprised this evaluates this way.
– Nathan
36 mins ago
It's set by thefor
loop. Each time through, it gets set to the next element of the list. What's the first element? Right, it's 0. What's the second? Right again, it's 1. etc.
– Tom Karzes
36 mins ago
1
"The for loop is a bit unusual in that it is using this element as the loop variable." - This may be where the confusion is coming from. Python isn't using the elementa[-1]
evaluates to, rather, it's using the indexa[-1]
itself. If you rephrase that statement, your answer becomes much more clear.
– Christian Dean
3 mins ago
|
show 1 more comment
How does it actuallygets set to 0, then 1, then 2 and finally 2
? That's the confusing part!
– Amir A. Shabani
40 mins ago
1
Agree, I don't really understand by reading this answer
– gameon67
39 mins ago
I expected it would be like sayingfor 3 in a: print(a[-1])
(becausea[-1]
was3
at the start of the loop) and give an error, but clearly that's not the case. You're obviously correct about what's happening, but I'm surprised this evaluates this way.
– Nathan
36 mins ago
It's set by thefor
loop. Each time through, it gets set to the next element of the list. What's the first element? Right, it's 0. What's the second? Right again, it's 1. etc.
– Tom Karzes
36 mins ago
1
"The for loop is a bit unusual in that it is using this element as the loop variable." - This may be where the confusion is coming from. Python isn't using the elementa[-1]
evaluates to, rather, it's using the indexa[-1]
itself. If you rephrase that statement, your answer becomes much more clear.
– Christian Dean
3 mins ago
How does it actually
gets set to 0, then 1, then 2 and finally 2
? That's the confusing part!– Amir A. Shabani
40 mins ago
How does it actually
gets set to 0, then 1, then 2 and finally 2
? That's the confusing part!– Amir A. Shabani
40 mins ago
1
1
Agree, I don't really understand by reading this answer
– gameon67
39 mins ago
Agree, I don't really understand by reading this answer
– gameon67
39 mins ago
I expected it would be like saying
for 3 in a: print(a[-1])
(because a[-1]
was 3
at the start of the loop) and give an error, but clearly that's not the case. You're obviously correct about what's happening, but I'm surprised this evaluates this way.– Nathan
36 mins ago
I expected it would be like saying
for 3 in a: print(a[-1])
(because a[-1]
was 3
at the start of the loop) and give an error, but clearly that's not the case. You're obviously correct about what's happening, but I'm surprised this evaluates this way.– Nathan
36 mins ago
It's set by the
for
loop. Each time through, it gets set to the next element of the list. What's the first element? Right, it's 0. What's the second? Right again, it's 1. etc.– Tom Karzes
36 mins ago
It's set by the
for
loop. Each time through, it gets set to the next element of the list. What's the first element? Right, it's 0. What's the second? Right again, it's 1. etc.– Tom Karzes
36 mins ago
1
1
"The for loop is a bit unusual in that it is using this element as the loop variable." - This may be where the confusion is coming from. Python isn't using the element
a[-1]
evaluates to, rather, it's using the index a[-1]
itself. If you rephrase that statement, your answer becomes much more clear.– Christian Dean
3 mins ago
"The for loop is a bit unusual in that it is using this element as the loop variable." - This may be where the confusion is coming from. Python isn't using the element
a[-1]
evaluates to, rather, it's using the index a[-1]
itself. If you rephrase that statement, your answer becomes much more clear.– Christian Dean
3 mins ago
|
show 1 more comment
The left expression of a for
loop statement gets assigned with each item in the iterable on the right in each iteration, so
for n in a:
print(n)
is just a fancy way of doing:
for i in range(len(a)):
n = a[i]
print(n)
Likewise,
for a[-1] in a:
print(a[-1])
is just a fancy way of doing:
for i in range(len(a)):
a[-1] = a[i]
print(a[-1])
where in each iteration, the last item of a
gets assigned with the next item in a
, so when the iteration finally comes to the last item, its value got last assigned with the second-last item, 2
.
add a comment |
The left expression of a for
loop statement gets assigned with each item in the iterable on the right in each iteration, so
for n in a:
print(n)
is just a fancy way of doing:
for i in range(len(a)):
n = a[i]
print(n)
Likewise,
for a[-1] in a:
print(a[-1])
is just a fancy way of doing:
for i in range(len(a)):
a[-1] = a[i]
print(a[-1])
where in each iteration, the last item of a
gets assigned with the next item in a
, so when the iteration finally comes to the last item, its value got last assigned with the second-last item, 2
.
add a comment |
The left expression of a for
loop statement gets assigned with each item in the iterable on the right in each iteration, so
for n in a:
print(n)
is just a fancy way of doing:
for i in range(len(a)):
n = a[i]
print(n)
Likewise,
for a[-1] in a:
print(a[-1])
is just a fancy way of doing:
for i in range(len(a)):
a[-1] = a[i]
print(a[-1])
where in each iteration, the last item of a
gets assigned with the next item in a
, so when the iteration finally comes to the last item, its value got last assigned with the second-last item, 2
.
The left expression of a for
loop statement gets assigned with each item in the iterable on the right in each iteration, so
for n in a:
print(n)
is just a fancy way of doing:
for i in range(len(a)):
n = a[i]
print(n)
Likewise,
for a[-1] in a:
print(a[-1])
is just a fancy way of doing:
for i in range(len(a)):
a[-1] = a[i]
print(a[-1])
where in each iteration, the last item of a
gets assigned with the next item in a
, so when the iteration finally comes to the last item, its value got last assigned with the second-last item, 2
.
answered 24 mins ago
blhsingblhsing
43.5k41744
43.5k41744
add a comment |
add a comment |
Kundan Verma is a new contributor. Be nice, and check out our Code of Conduct.
Kundan Verma is a new contributor. Be nice, and check out our Code of Conduct.
Kundan Verma is a new contributor. Be nice, and check out our Code of Conduct.
Kundan Verma is a new contributor. Be nice, and check out our Code of Conduct.
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6
Somehow this question looks like a bad and newbie question, but I don't get the logic either lol
– gameon67
38 mins ago
3
I don't know why you would ever want to do this, but now I know you can
– Nathan
35 mins ago
over the iteration only the last time a[-1] gets its value, all other times python treats as iteration variable.
– Arun Augustine
32 mins ago
6
This would be a great question for an awful coding interview
– Nathan
29 mins ago
@Nathan same idea, would be a great trivial question in interview
– gameon67
28 mins ago