用于检查字符串开头和结尾的空格的正则表达式

Question

This code supposed not to let strings with whitespaces in the beginning and end. In some reason I have negative result with this code

import re
def is_match(pattern, string):
    return True if len(re.compile(pattern).findall(string)) == 1 else False
print(is_match("[^\s]+[a-zA-Z0-9]+[^\s]+", '1'))

However, other strings work fine. Can anyone explain why result is negative, or even provide better function (newbie in python).

Answer 1

检查字符串开头或结尾是否有空格的最简单方法不涉及正则表达式。

if test_string != test_string.strip():

Answer 2

The regexp you're looking for is ^\\s|\\s$ :

xs = ["no spaces", "  starts", "ends  ", "\t\tboth\n\n", "okay"]

import re
print [x for x in xs if re.search(r'^\s|\s$', x)]

## ['  starts', 'ends  ', '\t\tboth\n\n']

^\\s.*?\\s$ only matches whitespace on both ends:

print [x for x in xs if re.search(r'^\s.*?\s$', x, re.S)]

## ['\t\tboth\n\n']

An inverse expression (no starting-ending whitespace) is ^\\S.*?\\S$ :

print [x for x in xs if re.search(r'^\S.*?\S$', x, re.S)]

## ['no spaces', 'okay']

Answer 3

def is_whiteSpace(string):
    t=' ','\t','\n','\r'
    return string.startswith(t) or string.endswith(t)


print is_whiteSpace(" GO") -> True
print is_whiteSpace("GO") -> False
print is_whiteSpace("GO ") -> True
print is_whiteSpace(" GO ") -> True

Answer 4

No fancy regex needed, just use the way more readable:

>>> def is_whitespace(s):
    from string import whitespace
    return any((s[0] in whitespace, s[-1] in whitespace))

>>> map(is_whitespace, ("foo", "bar ", " baz", "\tspam\n"))
[False, True, True, True]

Answer 5

Instead of trying to construct a regular expression that detects strings without spaces, it's easier to check for strings that DO have spaces and then invert the logic in your code.

Remember that re.match() returns None (a logical false value) if it doesn't find a match, and a SRE_Match object (logical true value) if it does find a match. Use that to write something like this:

In [24]: spaces_pattern = re.compile ( r"^(\s.+|.+\s)$" )

In [27]: for s in ["Alpha", " Bravo", "Charlie ", " Delta "]:
   ....:     if spaces_pattern.match(s):
   ....:         print ( "%s had whitespace." % s )
   ....:     else:
   ....:         print ( "%s did not have whitespace." % s )
   ....: 
Alpha did not have whitespace.
 Bravo had whitespace.
Charlie  had whitespace.
 Delta  had whitespace.

Note the use of the ^$ anchors to force the match over the entire input string.

---

Edit:

This doesn't even need regexp at all - you only need to check the first and last characters:

test_strings = ['a', ' b', 'c ', ' d ', 'e f', ' g h', ' i j', ' k l ']
for s in test_strings:
    if s[0] in " \n\r\t":
        print("'%s' started with whitespace." % s)
    elif s[-1] in " \n\r\t":
        print("'%s' ended with whitespace." % s)
    else:
        print("'%s' was whitespace-free." % s)

---

Edit 2:

A regex that should work anywhere: ^\\S(.*\\S)? . You may need to come up with a local equivalent to \\S ("anything but whitespace") if your regex dialect doesn't include it.

test_strings = ['a', ' b', 'c ', ' d ', 'e f', ' g h', ' i j', ' k l ']
import re

pat = re.compile("^\S(.*\S)?$")

for s in test_strings:
    if pat.match(s):
        print("'%s' had no whitespace." % s)
    else:
        print("'%s' had whitespace." % s)

Note that \\S is the negated form of \\s , ie \\S means "anything but whitespace".

Also note that strings of length 1 are accounted for by making part of the match optional. (You might think to use \\S.*\\S , but this forces a match length of at least 2.)

'a' had no whitespace.
' b' had whitespace.
'c ' had whitespace.
' d ' had whitespace.
'e f' had no whitespace.
' g h' had whitespace.
' i j' had whitespace.
' k l ' had whitespace.

Answer 6

A variant on ch3ka's suggestion:

import string
whitespace = tuple(string.whitespace)

'a '.endswith(whitespace)
## True

'a '.startswith(whitespace)
## False

'a\n'.endswith(whitespace)
## True

'a\t'.endswith(whitespace)
## True

I find it easier to remember than the regex stuff (except maybe the bit converting whitespace to a tuple).