I'm trying to match all patterns that end in bar. This is my regex pattern ".*bar$"
. I get no result... same thing happens if I use the carrot in to match at the beginning of patterns.
string = """
foo bar baz
bar foo baz
baz foo bar
bar baz foo
foo baz bar
baz bar foo
"""
search = re.findall(".*bar$", string)
for i in search:
print(i)
Your pattern works if you set the re.MULTILINE
flag. That way your pattern is matched on a line-by-line basis, so the $
matches line endings in addition to the ending of the string as a whole.
# Result: ['baz foo bar', 'foo baz bar']
search = re.findall(".*bar$", string, flags=re.MULTILINE)
Edit: Looks like you just want everything that ends in bar
, regardless of line endings. In that case, you can tell set the star *
to be non-greedy by adding a ?
:
>>> re.findall(".*?bar", "danibarsambarbreadbar")
['danibar', 'sambar', 'breadbar']
You can try this
import re
a ="foo bar baz\nbar foo baz\nbaz foo bar\nbar baz foo\nfoo baz bar\nbaz bar foo"
search = re.finditer("(.+bar)\n", a)
for i in search:
print(i.group())
Output:
baz foo bar
foo baz bar
Or you can try This:
import re
a ="foo bar baz\nbar foo baz\nbaz foo bar\nbar baz foo\nfoo baz bar\nbaz bar foo"
search = re.findall("(.+bar)\n", a)
print(search)
Output:
['baz foo bar', 'foo baz bar']
Try this:
string = """
foo bar baz
bar foo baz
baz foo bar
bar baz foo
foo baz bar
baz bar foo
"""
re.findall(r"(.+bar)\n", string)
output:
['baz foo bar', 'foo baz bar']
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