Replacing slice by an iterable-CodePudding

I have a list s which looks as below:

s = [1, 2, 3]

I am replacing slice of 's' using below code:

s[1:4] = [22, 3, 4, 5, 6, 7]
print(s)

Output: [1, 22, 3, 4, 5, 6, 7]

My understanding s[1:4] should replace only 3 elements starting from 1st element and up to but not including 4th element.

Assumed Output: [1, 22, 3, 4]

CodePudding user response：

So let's use a better example. I've included the output as commented out code in the code snippets

s = list(range(1, 11))
print(f"Len = {len(s)} | S = {s}")

# Len = 10 | S = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

There are 3 scenarios:

Case 1: Assigned list is shorter than the index range

s = list(range(1, 11))
s[1:4] = [0, 0]
print(f"Len = {len(s)} | S = {s}")

# Len = 9 | S = [1, 0, 0, 5, 6, 7, 8, 9, 10]

Ans : The list size is shrunk by 1 since only 2 items are being assigned to 3 places.

Case 2: Assigned list is longer than the index range

s = list(range(1, 11))
s[1:4] = [0, 0, 0, 0, 0, 0, 0]
print(f"Len = {len(s)} | S = {s}")

# Len = 14 | S = [1, 0, 0, 0, 0, 0, 0, 0, 5, 6, 7, 8, 9, 10]

Ans : The list size is increased by 4 since 7 items are being assigned to 3 places.

Case 3: Assigned list is equal to index range

s = list(range(1, 11))
s[1:4] = [0, 0, 0]
print(f"Len = {len(s)} | S = {s}")

# Len = 10 | S = [1, 0, 0, 0, 5, 6, 7, 8, 9, 10]

Ans : In this case it will be replaced properly since both the sizes match.

Conclusion

Python seems to be shrinking and growing the list to accommodate the assigned items.
All elements outside the slice indices will not be affected regardless of how big/small the assigned list is.

Update

As @mealhour pointed out there is a fourth case where you can increase the step size to be greater than 1 and assign it to every kth item in the list. In this case, the sizes have to match otherwise, python throws an error. This StackOverflow question explains it really well

s = list(range(1, 11))
# s[1:4:2] = [0, 0, 0] <- This throws an error
s[1:4:2] = [0, 0]
print(f"Len = {len(s)} | S = {s}")

# Len = 10 | S = [1, 0, 3, 0, 5, 6, 7, 8, 9, 10]

CodePudding user response：

s = [1,2,3]
x = [22,3,4,5,6]
j = 0
for i in range(1,4):
    try:
        s[i] = x[j]
    except IndexError:
        s.append(x[j])
    j  =1

You can use something like this to achieve your goal