Decompose a Python string into its characters

I want to break a Python string into its characters.

sequenceOfAlphabets = list( string.uppercase )

works.

However, why does not

sequenceOfAlphabets = re.split( '.', strin开发者_Python百科g.uppercase )

work?

All I get are empty, albeit expected count of elements

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The '.' matches every character and re.split returns everything that wasn't matched, that's why you're getting the empty list.

Using list is usually the way to handle something like this but if you want to use regular expressions just use re.findall

sequenceOfAlphabets = re.findall( '.', string.uppercase )

That should give you ['A', 'B', 'C', .... ,'Z']

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Because the delimiter character used by split does not appear in the resulting list. This allows it be used like:

re.split(',', "foo,bar,baz")
['foo', 'bar', 'baz']

Also, you will find the resulting list from your split code actually contains one extra element, since split returns one more than the number of delimiters found. The above has two commas, so it returns a three-element list.

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If you can do something with both a built-in function and with regexes, then usually the built-in approach will be faster and more legible.

The regex world is a maze of twisty little passages, populated by purveyors of almost-truths like """The '.' matches every character""" ... which it does, but only when you use the re.DOTALL flag. This information is not cunningly concealed in the fine print of the documentation; it's right there as the FIRST entry of "special characters":

'.'
(Dot.) In the default mode, this matches any character except a newline. If the DOTALL flag has been specified, this matches any character including a newline.

>>> import re
>>> re.findall(".", "fu\nbar")
['f', 'u', 'b', 'a', 'r']
>>>

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Just an FYI, this also works:

sequenceOfAlphabets = [a for a in string.uppercase]

...but that does exactly what list() would do so I don't think it would be any faster (I could be wrong).

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You can also create an empty set and use the update method, like so:

destroy_string = set()
destroy_string.update('Stack Overflow')
destroy_string
{'k', ' ', 'S', 'c', 'v', 'o', 'r', 't', 'w', 'e', 'f', 'O', 'l', 'a'}

Albeit, it will become unordered and the duplicates will be lost in the set, however, this is still a valid way to decompose a string into a set of its individual members.

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From the documentation:

If capturing parentheses are used in pattern, then the text of all groups in the pattern are also returned as part of the resulting list.

Also note:

If there are capturing groups in the separator and it matches at the start of the string, the result will start with an empty string. The same holds for the end of the string.

So, use re.split( '(.)', string.uppercase)[1:-1] instead.