1

This is my current snippet:

l = ['i', 'b', 'c', 'a', '3', 'a', '7', 'a', '9', 'k', 'a', '66', 'k', 'd']

new_list = []
a = {}
for i in range(len(l)):
    if 'a' == l[i]:
        a['plus_line'] = l[i+1]
        new_list.append(a)


print(new_list)

I am expecting output like this below:

[{'plus_line': '3'}, {'plus_line': '7'}, {'plus_line': '9'}, {'plus_line': '66'}]

But I am getting:

[{'plus_line': '66'}, {'plus_line': '66'}, {'plus_line': '66'}, {'plus_line': '66'}]

I am trying to achieve when a will be in the list, then the next item of the a will be the value of the dict.

Improve this question
3
  • 7
    your expected and actual output are the same Commented Oct 28, 2019 at 15:10
  • 1
    Hi sactual output is: [{'plus_line': '66'}, {'plus_line': '66'}, {'plus_line': '66'}, {'plus_line': '66'}] Commented Oct 28, 2019 at 15:14
  • 1
    a = [{'plus_line': y} for x, y in zip(l, l[1:]) if x == 'a']. Commented Oct 28, 2019 at 15:16

4 Answers 4

Reset to default
4

Your problem is that you're mutating the dict named a. Instead you could do something like:

l = ['i', 'b', 'c', 'a', '3', 'a', '7', 'a', '9', 'k', 'a', '66', 'k', 'd']

new_list = []
for i in range(len(l)):
    if 'a' == l[i]:
        new_list.append({'plus_line': l[i+1]})

print(new_list)

This will create a new dict instead of mutating an existing one.

Improve this answer
Sign up to request clarification or add additional context in comments.

Comments

2

itzhaki explains the problem with your code, here's another possible way of doing it:

l = ['i', 'b', 'c', 'a', '3', 'a', '7', 'a', '9', 'k', 'a', '66', 'k', 'd']

result = [{'plus_line':l[n+1]} for n, i in enumerate(l) if i == 'a']

print(result)

Output:

[{'plus_line': '3'},
 {'plus_line': '7'},
 {'plus_line': '9'},
 {'plus_line': '66'}]
Improve this answer

1 Comment

kinda weird to accept an answer that shows an alternative implementation without explaining the error but what do I know :(
0

You simply needed to reset your a = {} at the start of each loop (not outside the loop). By initializing it outside the loop, you created one dictionary named a, which has only one key, plus line which keeps getting assigned a new value. 

l = ['i', 'b', 'c', 'a', '3', 'a', '7', 'a', '9', 'k', 'a', '66', 'k', 'd']

new_list = []
for i in range(len(l)):
    a = {}
    if 'a' == l[i]:
        a['plus_line'] = l[i+1]
        new_list.append(a)


print(new_list)

OUTPUT:

[{'plus_line': '3'}, {'plus_line': '7'}, {'plus_line': '9'}, {'plus_line': '66'}]
Improve this answer

Comments

0

See this:

>>> d = {}
>>> d['hi there'] = 23
>>> d['hi there'] = 24
>>> d
{'hi there': 24}

You're overwriting the key 'plus_line' of dictionary a.

Items you're appending in list new_list are referencing to dictionary a. Thus the problem.

Possible solution for that can be:

>>> new_list = []
>>> for i in range(len(l)):
...     if 'a' == l[i] and i < len(l) - 2:
...             new_list.append({'plus_line':l[i+1]})
... 
>>> new_list
[{'plus_line': '3'}, {'plus_line': '7'}, {'plus_line': '9'}, {'plus_line': '66'}]
Improve this answer

1 Comment

Maybe just limit the for loop to for i in range(len(l)-1): ?