Not asking particularly about the slicing, Basically I want to replace the value of specific index of list by 2 or more elements.
For Example:
list_a = [1, 2, 3] and list_b = [4, 5] so what I want is result = [4, 5, 2, 3]
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harry jame
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Are you trying to remove an element from the list at a specific index and introduce a new list of elements in that position? – thefourtheye Dec 09 '22 at 08:32
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Not a list, just elements of the list – harry jame Dec 09 '22 at 08:32
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Simply `result = list_b + list_a[1:]`…?! – deceze Dec 09 '22 at 08:43
2 Answers
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If you want a new list:
idx = 0
result = list_a[:idx]+list_b+list_a[idx+1:]
Output: [4, 5, 2, 3]
If you want to modify the list in place:
idx = 0
list_a[:] = list_a[:idx]+list_b+list_a[idx+1:]
# OR (see @Nineteendo's comment)
list_a[idx:idx + 1] = list_b
step by step
# get values up to the insertion point
list_a[:idx]
# []
# get values after the insertion point
list_a[idx+1:]
# [2, 3]
# combine everything in a single list
list_a[:idx]+list_b+list_a[idx+1:]
# [4, 5, 2, 3]
mozway
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Can you explain a little bit more as I am very new to python, kind of noob, you can say – harry jame Dec 09 '22 at 08:37
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@mozway, can't you also use `list_a[idx:idx + 1] = list_b` for the second solution? – Nineteendo Dec 09 '22 at 08:41
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Yeah, now I can understand that easily, like a string slicing. Thanks – harry jame Dec 09 '22 at 08:42
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You can achieve this using below,
First you need to pop 0 index value, then try to reverse the b and loop over to insert values of b in 0 index of a
a = [1, 2, 3]
b = [4, 5]
a.pop(0)
for i in b.reverse():
a.insert(0, i)
print(a)
Output:
[4, 5, 2, 3]
Usman Arshad
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It's a nice one (I also had this in mind), but not efficient if you have a large `list_b` as `insert` is O(n) – mozway Dec 09 '22 at 08:37
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@mozway - Right! Just wanted to give the understanding how we can achieve in simpler way – Usman Arshad Dec 09 '22 at 08:39
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Yeah, this solution is self-explainable, I understand it clearly, Thanks Usman Arshad – harry jame Dec 09 '22 at 08:39