[英]python replace regex match without spaces
I basically want to 'join' numbers that should clearly go together.我基本上想“加入”应该清楚地在一起的数字。 I want to replace the regex match with itself but without any spaces.
我想用它自己替换正则表达式匹配,但没有任何空格。
I have:我有:
df
a
'Fraxiparine 9 500 IU (anti-Xa)/1 ml'
'Colobreathe 1 662 500 IU inhalačný prášok v tvrdej kapsule'
I want to have:我希望有:
df
a
'Fraxiparine 9500 IU (anti-Xa)/1 ml'
'Colobreathe 1662500 IU inhalačný prášok v tvrdej kapsule'
I'm using r'\d+\s+\d+\s*\d+'
to match the numbers, and I've created the following function to remove the spaces within the string:我正在使用
r'\d+\s+\d+\s*\d+'
来匹配数字,并且我创建了以下函数来删除字符串中的空格:
def spaces(x):
match = re.findall(r'\d+\s+\d+\s*\d+', x)
return match.replace(" ","")
Now I'm having trouble applying that function to the full dataframe, but I also don't know exactly how to replace the original match with the string without any spaces.现在我无法将该函数应用于完整的数据帧,但我也不知道如何用没有任何空格的字符串替换原始匹配。
Try using the following code:尝试使用以下代码:
def spaces(s):
return re.sub('(?<=\d) (?=\d)', '', s)
df['a'] = df['a'].apply(spaces)
The regex will match:正则表达式将匹配:
(?<=\d)
(?<=\d)
(?=\d)
.(?=\d)
。 Then, the pandas.Series.apply function will apply your function to all rows of your dataframe.然后, pandas.Series.apply函数会将您的函数应用于数据框的所有行。
Output:输出:
0 Fraxiparine 9500 IU (anti-Xa)/1 ml
1 Colobreathe 1662500 IU inhalačný prášok v tvrd...
I believe that your problem can be solved by tweaking a bit your function in order to be applied on the whole string 'match' as follows :我相信您的问题可以通过稍微调整您的函数来解决,以便应用于整个字符串“匹配”,如下所示:
import pandas as pd
import re
df = pd.DataFrame({'a' : ['Fraxiparine 9 500 IU (anti-Xa)/1 ml','Colobreathe 1 662 500 IU inhalačný prášok v tvrdej kapsule']})
# your function
def spaces(x):
match = re.findall(r'\d+\s+\d+\s*\d+', x)
replace_with = match[0].replace(" ","")
return x.replace(match[0], replace_with)
# now apply it on the whole dataframe, row per row
df['a'] = df['a'].apply(lambda x: spaces(x))
Use利用
df['a'] = df['a'].str.replace(r'(?<=\d)\s+(?=\d)', '', regex=True)
EXPLANATION解释
NODE EXPLANATION
--------------------------------------------------------------------------------
(?<= look behind to see if there is:
--------------------------------------------------------------------------------
\d digits (0-9)
--------------------------------------------------------------------------------
) end of look-behind
--------------------------------------------------------------------------------
\s+ whitespace (\n, \r, \t, \f, and " ") (1 or
more times (matching the most amount
possible))
--------------------------------------------------------------------------------
(?= look ahead to see if there is:
--------------------------------------------------------------------------------
\d digits (0-9)
--------------------------------------------------------------------------------
) end of look-ahead
If your plan is to remove spaces only in \d+\s+\d+\s*\d+
:如果您的计划是仅删除
\d+\s+\d+\s*\d+
中的空格:
df['a'] = df['a'].str.replace(r'\d+\s+\d+\s*\d+', lambda m: re.sub(r'\s+', '', m.group()), regex=True)
See str.replace
:见
str.replace
:
repl : str or callable
repl : str 或可调用
Replacement string or a callable.替换字符串或可调用对象。 The callable is passed the regex match object and must return a replacement string to be used.
可调用对象传递正则表达式匹配对象,并且必须返回要使用的替换字符串。 See re.sub().
参见 re.sub()。
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