具有[：numeric：]的Python正则表达式

Question

I am having some trouble with Python giving me a result I do not expect. Here is a sample code :

number = re.search(" [0-9] ", "test test2 test_ 2 333")
print number.groups()

number = re.search(" [[:digit:]] ", "test test2 test_ 2 333")
print number.groups()

In the first block I get an object returned but with nothing in it. Where I think I should get the string "2".

In the second block I don't even get an object, where I am expection the string "2".

While when I do this in bash everything looks fine :

echo "test test2 test_ 2 333" | grep " [[:digit:]] "
echo "test test2 test_ 2 333" | grep " [0-9] "

Can somebody help me please?

Answer 1

The groups() method returns the capture groups. It does not return group 0, in case that's what you were expecting. Use parens to indicate capture groups. eg:

>>> number = re.search(" ([0-9]) ", "test test2 test_ 2 333")
>>> print number.groups()
('2',)

For your second example, Python's re module doesn't recognize the "[:digit:]" syntax. Use \\d . eg:

>>> number = re.search(r" (\d) ", "test test2 test_ 2 333")
>>> print number.groups()
('2',)

Answer 2

You are missing the () which capture the contents for use with the groups() (and other) function(s).

number = re.search(" ([0-9]) ", "test test2 test_ 2 333")
print number.groups()

This however won't work because python does not support the [[:number:]] notation

number = re.search(" ([[:digit:]]) ", "test test2 test_ 2 333")
print number.groups()

Answer 3

Is this what you're looking for?

>>> re.findall(r'([0-9])', "test test2 test_ 2 333")
['2', '2', '3', '3', '3']

Answer 4

number = re.search(" [0-9] ", "test test2 test_ 2 333")
print number.group(0)

groups() only returns groups 1 and up (a bit odd if you're used to other languages).

Answer 5

.groups() returns values inside of matched parentheses. This regex doesn't have any regions defined by parens so groups returns nothing. You want:

m = re.search(" ([0-9]) ", "test test2 test_ 2 333") m.groups() ('2',)