I'm looking for the most pythonic way of doing something like that:
a = [1,2,3,4,5,6,7,8]
b = ['a','b','c']
c = replace(a,b,2)
c is now [1,2,'a','b','c',6,7,8]
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you need to replace the part of array from index 2? that is the third parameter?Keerthana Prabhakaran– Keerthana Prabhakaran2017-04-02 14:24:17 +00:00Commented Apr 2, 2017 at 14:24
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What's your guess? have you searched for proper functions or methods that might be useful for this task?Kasravnd– Kasravnd2017-04-02 14:24:27 +00:00Commented Apr 2, 2017 at 14:24
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2 Answers
you can slice your list accordingly!
That is, with
aandbbeing your initial list and the one that you want to replace from indexs,a[:s]will get all elements before from 0 to s that is ([1,2]).a[s+len(b):]will get all items from indexstolen(b), that is ([6,7,8])
so when you concatenate the first result along with b and then the second result you can get the desired output!
a[:s]+b+a[s+len(b):]
So,
>>> a = [1,2,3,4,5,6,7,8]
>>> b = ['a','b','c']
>>> replace=lambda a,b,s:a[:s]+b+a[s+len(b):]
>>> c = replace(a,b,2)
>>> print c
[1, 2, 'a', 'b', 'c', 6, 7, 8]
Hope this helps!
answered Apr 2, 2017 at 14:26
Keerthana Prabhakaran
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1 Comment
Mohideen bin Mohammed
pythonic answer :)
Okay, here is another simple way to do this: (Edit: forgot slicing a to copy it into c)
a = [1,2,3,4,5,6,7,8]
b = ['a','b','c']
c=a[:]
c[2:5]=b
print(c)
Or, when you want to replace in place:
a = [1,2,3,4,5,6,7,8]
b = ['a','b','c']
a[2:5]=b
print(a)
Hope I helped!
5 Comments
Keerthana Prabhakaran
this replace the original list!! that is even
a is affected!Keerthana Prabhakaran
@Keiwan +1 for using
a=a[:]
trashy
Yeah, I just noticed and edited. Thank you @Keiwan
Virgo
That is a very tricky python feature in my opinion!
JDune
This is really cool, and is simpler than the accepted answer. Thanks!