How get the value of counter in python for loop

Question

What is the best way to fix the following code

my_list=[]
elem={}
for i in range(4):
    elem['id']=i
    my_list.append(elem)
print my_list

result

[{'id': 3}, {'id': 3}, {'id': 3}, {'id': 3}]

expected result

[{'id': 0}, {'id': 1}, {'id': 2}, {'id': 3}]

**I don't want to use another variable

Hackaholic · Accepted Answer · 2015-04-04 02:36:24Z

4

Pythonic way: list comprehension

>>> [{'id':x} for x in range(4)]
[{'id': 0}, {'id': 1}, {'id': 2}, {'id': 3}]

answered Apr 4, 2015 at 2:36

Hackaholic

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Comments

Avinash Raj · Accepted Answer · 2015-04-04 02:41:41Z

4

You need to empty the dictionary inside the for loop itself.

my_list=[]
elem={}
for i in range(4):
    elem['id']=i
    my_list.append(elem)
    elem={}
print my_list

Output:

[{'id': 0}, {'id': 1}, {'id': 2}, {'id': 3}]

OR

simply,

my_list=[]
for i in range(4):
    elem = {}
    elem['id']=i
    my_list.append(elem)

print my_list

edited Apr 4, 2015 at 2:41

answered Apr 4, 2015 at 2:35

Avinash Raj

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4 Comments

Tanveer Alam Over a year ago

for i in range(4): elem={} elem['id']=i my_list.append(elem) I think we don't need to declare elem={} twice.

Tarik Over a year ago

We can also move elem={} as the first statement within the loop and get rid of the first one right before the loop.

Tarik Over a year ago

@TanveerAlam elem is never declared twice during execution. It's just that the code can be shrunk down and be better structured.

Tanveer Alam Over a year ago

@Tarik I just went through your profile Sir and I think you have typo error in about me section. You have mentioned Haslell which should be Haskell I think.

Shashank · Accepted Answer · 2015-04-04 02:42:42Z

3

my_list = []
for i in range(4):
    mylist.append({'id': i})

This is practically equivalent to:

my_list = [{'id': i} for i in range(4)]

answered Apr 4, 2015 at 2:42

Shashank

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