Why doesn’t the slice’s step index work with ‘replace’?

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February 11th, 2013, 03:09 AM

Akshat1

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Why doesn’t the slice’s step index work with ‘replace’?

Here’s my code:
>>> a = ‘Hello Bello Jello!’
>>> a = a.replace(a[::2],’H’)
>>> a
'Hello Bello Jello!'
>>> a = a.replace(a[0:2],'H')
>>> a
'Hllo Bello Jello!'
While using the ‘replace’ command; without the step index in slice, it works but when I include the step index, it doesn't and returns the previously assigned value of ‘a’. Why is that?

February 11th, 2013, 04:55 AM

SuperOscar

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Quote:

Originally Posted by Akshat1

While using the ‘replace’ command; without the step index in slice, it works but when I include the step index, it doesn't and returns the previously assigned value of ‘a’. Why is that?

Simple. a[::2] equals to “HloBloJlo” in your example: every second letter in the string “a”. When you do:

Code:

a = a.replace(a[::2], 'H')

you are asking to replace the string “HloBloJlo” with “H” – but since “HloBloJlo” doesn’t appear in a, it cannot be replaced! So the string is returned intact.

OTOH, in

Code:

a = a.replace(a[0:2], 'H')

you are asking to replace “He” (= a[0:2]) with “H”, and this string is a substring of “a” and can be replaced.

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February 11th, 2013, 10:31 AM

b49P23TIvg

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Solution to the step replace

Code:

>>> a = 'Hello Bello Jello!'
>>> la = list(a)
>>> la
['H', 'e', 'l', 'l', 'o', ' ', 'B', 'e', 'l', 'l', 'o', ' ', 'J', 'e', 'l', 'l', 'o', '!']
>>> la[::2] = 'H'*(len(la[::2]))
>>> la
['H', 'e', 'H', 'l', 'H', ' ', 'H', 'e', 'H', 'l', 'H', ' ', 'H', 'e', 'H', 'l', 'H', '!']
>>> ''.join(la)
'HeHlH HeHlH HeHlH!'
>>>

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Last edited by b49P23TIvg : February 11th, 2013 at 10:33 AM.

February 14th, 2013, 07:58 AM

Akshat1

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Quote:

Originally Posted by SuperOscar

Simple. a[::2] equals to “HloBloJlo” in your example: every second letter in the string “a”. When you do:

Code:

a = a.replace(a[::2], 'H')

you are asking to replace the string “HloBloJlo” with “H” – but since “HloBloJlo” doesn’t appear in a, it cannot be replaced! So the string is returned intact.

OTOH, in

Code:

a = a.replace(a[0:2], 'H')

you are asking to replace “He” (= a[0:2]) with “H”, and this string is a substring of “a” and can be replaced.

Thank you soooo much. Thanks

Why doesn’t the slice’s step index work with ‘replace’?

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