从pandas dataframe中删除包含字符串中数字值的行

Question

I have a pandas data frame with 2 columns, type and text The text column contains string values. How can I delete rows which contains some numeric values in the text column. eg:

`ABC 1.3.2`, `ABC12`, `2.2.3`, `ABC 12 1`

I have tried below, but get an error. Any idea why this is giving error?

df.drop(df[bool(re.match('^(?=.*[0-9]$)', df['text'].str))].index)

Answer 1

In your case, I think it's better to use simple indexing rather than drop. For example:

>>> df
       text type
0       abc    b
1    abc123    a
2       cde    a
3  abc1.2.3    b
4     1.2.3    a
5       xyz    a
6    abc123    a
7      9999    a
8     5text    a
9      text    a


>>> df[~df.text.str.contains(r'[0-9]')]
   text type
0   abc    b
2   cde    a
5   xyz    a
9  text    a

That locates any rows with no numeric text

To explain:

df.text.str.contains(r'[0-9]')

returns a boolean series of where there are any digits:

0    False
1     True
2    False
3     True
4     True
5    False
6     True
7     True
8     True
9    False

and you can use this with the ~ to index your dataframe wherever that returns false

Answer 2

Data from jpp

s[s.str.isalpha()]
Out[261]: 
0    ABC
2    DEF
6    GHI
dtype: object

Answer 3

Assuming you define numeric as x.isdigit() evaluating to True , you can use any with a generator expression and create a Boolean mask via pd.Series.apply :

s = pd.Series(['ABC', 'ABC 1.3.2', 'DEF', 'ABC12', '2.2.3', 'ABC 12 1', 'GHI'])

mask = s.apply(lambda x: not any(i.isdigit() for i in x))

print(s[mask])

0    ABC
2    DEF
6    GHI
dtype: object

Answer 4

Well as I asked in the comment, what is your defintion of numeric. If we follow python's isnumeric with split() we get the following:

import pandas as pd

import pandas as pd

df = pd.DataFrame({
    'col1': ['ABC', 'ABC 1.3.2', 'DEF', 'ABC12', '2.2.3', 'ABC 12 1', 'GHI']
})

m1 = df['col1'].apply(lambda x: not any(i.isnumeric() for i in x.split()))
m2 = df['col1'].str.isalpha()
m3 = df['col1'].apply(lambda x: not any(i.isdigit() for i in x))
m4 = ~df['col1'].str.contains(r'[0-9]')

print(df.assign(hasnonumeric=m1,isalhpa=m2, isdigit=m3, contains=m4))

# Opting for hasnonumeric
df = df[m1]

prints:

        col1  hasnonumeric  isalhpa  isdigit  contains
0        ABC          True     True     True      True
1  ABC 1.3.2          True    False    False     False
2        DEF          True     True     True      True
3      ABC12          True    False    False     False
4      2.2.3          True    False    False     False
5   ABC 12 1         False    False    False     False
6        GHI          True     True     True      True