过滤'pandas'中不包含字母（alpha）的所有行
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Filter all rows that do not contain letters (alpha) in ′pandas′
提问by owwoow14
I am trying to filter a pandasdataframe using regular expressions.
I want to delete those rows that do not contain any letters. For example:
我正在尝试pandas使用regular expressions. 我想删除那些不包含任何字母的行。例如：
Col A.
50000
7848
dog
cat 583
rabbit 444
My desired results is:
我想要的结果是：
Col A.
dog
cat 583
rabbit 444
I have been trying to solve this problem unsuccessful with regexand pandasfilter options. See blow. I am specifically running into problems when I try to merge two conditions for the filter. How can I achieve this?
我一直在试图解决这个问题不成功regex和pandas过滤器选项。见打击。当我尝试合并过滤器的两个条件时，我特别遇到了问题。我怎样才能做到这一点？
Option 1:
选项1：
df['Col A.'] = ~df['Col A.'].filter(regex='\d+')
Option 2
选项 2
df['Col A.'] = df['Col A.'].filter(regex=\w+)
Option 3
选项 3
from string import digits, letters
df['Col A.'] = (df['Col A.'].filter(regex='|'.join(letters)))
OR
或者
df['Col A.'] = ~(df['Col A.'].filter(regex='|'.join(digits)))
OR
或者
df['Col A.'] = df[~(df['Col A.'].filter(regex='|'.join(digits))) & (df['Col A.'].filter(regex='|'.join(letters)))]
回答by jezrael
I think you'd need str.containsto filter values which contain letters by the means of boolean indexing:
我认为您需要str.contains通过以下方式过滤包含字母的值boolean indexing：
df = df[df['Col A.'].str.contains('[A-Za-z]')]
print (df)
Col A.
2 dog
3 cat 583
4 rabbit 444
If there are some NaNs values you can pass a parameter:
如果有一些NaNs 值，您可以传递一个参数：
df = df[df['Col A.'].str.contains('[A-Za-z]', na=False)]
print (df)
Col A.
3 dog
4 cat 583
5 rabbit 444
回答by Samuel GIFFARD
Have you tried:
你有没有尝试过：
df['Col A.'].filter(regex=r'\D') # Keeps only if there's a non-digit character
or:
或者：
df['Col A.'].filter(regex=r'[A-Za-z]') # Keeps only if there's a letter (alpha)
or:
或者：
df['Col A.'].filter(regex=r'[^\W\d_]') # More info in the link below...
Explanation: https://stackoverflow.com/a/2039476/8933502
回答by Vaidic
df['Col A.'].str.contains(r'^\d+$', na=True)# if string with only digits or if int/float then will result in NaN converted to True
df['Col A.'].str.contains(r'^\d+$', na=True)# 如果字符串只有数字或者如果是 int/float 则将导致 NaN 转换为 True
eg: [50000, '$927848', 'dog', 'cat 583', 'rabbit 444', '3 e 3', 'e 3', '33', '3 e'] will give : [True,False,False,False,False,False,False, True,False]
例如：[50000, '$927848', 'dog', 'cat 583', 'rabbit 444', '3 e 3', 'e 3', '33', '3 e'] 将给出： [True,False ,假,假,假,假,假,真,假]
回答by Beno?t Zu
You can use ^.*[a-zA-Z].*$
您可以使用 ^.*[a-zA-Z].*$
https://regex101.com/r/b84ji1/1
https://regex101.com/r/b84ji1/1
Details
细节
^: Start of the line
^: 行首
.*: Match any character
.*: 匹配任何字符
[a-zA-Z]: Match letters
[a-zA-Z]: 匹配字母
$: End of the line
$： 队伍的尽头
