python - Edit a list while iterating over it (the pythonic way!) -

so, know how this, not seem quite pythonic mee. there cleaner way of doing this?

arr = list(range(10)) print(arr) n in range(len(arr)):     # perform som operation on element changes value "in place"     arr[n] += 1 print(arr) 

output (which how want it):

[0, 1, 2, 3, 4, 5, 6, 7, 8, 9] [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] 

what said. if want operation while you're creating list can do:

arr = [u + 1 u in range(10)] 

i presume you're using python 3. if you're using python 2, change range xrange (unless list size small), since python 2 range function returns actual list, xrange returns iterator; python 3 range function returns iterator that's similar old xrange (but few improvements).

if arr existing list want modify without replacing current list object new list object , can do

arr[:] = [u + 1 u in arr] 

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here's code illustrates difference between using arr , arr[:] on left hand side of assignment.

arr = list(range(10)) b = arr print(id(arr), id(b)) arr = [u + 1 u in arr] print(id(arr), id(b)) print(arr, b)  c = arr print(id(arr), id(c)) arr[:] = [u + 1 u in arr] print(id(arr), id(c)) print(arr, c) 

output

3073636268 3073636268 3073629964 3073636268 [1, 2, 3, 4, 5, 6, 7, 8, 9, 10] [0, 1, 2, 3, 4, 5, 6, 7, 8, 9] 3073629964 3073629964 3073629964 3073629964 [2, 3, 4, 5, 6, 7, 8, 9, 10, 11] [2, 3, 4, 5, 6, 7, 8, 9, 10, 11] 

so arr = [u + 1 u in arr] binds name arr new list object, arr[:] = [u + 1 u in arr] mutates existing list object. "under hood" creates new temporary list object , copies contents old list object.