Question

所以我试图按单元格编辑整个列单元格，将列从包含整数和字符串的内容更改为整数组件。例如

一个单元格如下：

3001234; textTEXT TextTeXTExt.TExt

我正在使用此命令：

df2.columns[3] = df2.columns[3].map(lambda x: x.lstrip([5:]))

我也尝试过这样的事情：

df2.columns[3] = df2.columns[3].split([])

这是我从ipython获得的错误：

AttributeError: 'unicode' object has no attribute 'map'

数据框中的实际列：

0                           11212; xxxxxxxxxx xxxxxxxx   
1                           11212; xxxxxxxxxx xxxxxxxx   
2                           11212; xxxxxxxxxx xxxxxxxx   
3                           11212; xxxxxxxxxx xxxxxxxx     
8                  667788; xxxxxxx xxxxxxxxxxxxx xxxxxx   
9                  55555; xxxxxxx xxxxxxxxxxxxx xxxxxx   
10                 55555; xxxxxxx xxxxxxxxxxxxx xxxxxx   
11                 55555; xxxxxxx xxxxxxxxxxxxx xxxxxx   
12                 33333; xxxxxxx xxxxxxxxxxxxx xxxxxx   
13                 333; xxx xxxxx @ xxx xxx 2 xxxx   
14                 9991; xxxx; xxxxxx xxxxx xxxx @ 2 xxx   
18                       1635; vvvvvvvvvvvv vvvvvv 10   
19                       1635; vvvvvvvvvvvv vvvvvv 10   
20                       1635; vvvvvvvvvvvv vvvvvv 10   
21                       1635; vvvvvvvvvvvv vvvvvv 10     
32                       1712; Cxxxx xxxxxxxx; xxx 0   
33                       1712; Cxxxx xxxxxxxx; xxx 0   
34                       1712; Cxxxx xxxxxxxx; xxx 0   
35                       1712; Cxxxx xxxxxxxx; xxx 0

这是我正在运行的代码

    import pandas as pd 

    # import excel file 
    xlsx = pd.ExcelFile("/home/PATH") 
    # create data frame from excel file on sheet 1
    df2 = pd.read_excel(xlsx,'Sheet1')

    df = pd.DataFrame({"Card": df2})
    print(df.head())

    df.iloc[:,0] = df.iloc[:,0].apply(lambda x: x.split(';')[0])
    print df.head()

    # delete columns not relative to us
    df2.drop(df2.columns[[0,5,10,11]],inplace=True,axis=1)

Answer 1

如果我理解你的问题，你可以试试这个：

import pandas as pd
import re
df = pd.DataFrame({'col1':['3001234; textTEXT TextTeXTExt.TExt', '1005678;  more text']})
print(df)
col1
0  3001234; textTEXT TextTeXTExt.TExt
1                  1005678; more text


digits = df['col1'].apply(lambda x: re.findall('\d+', str(x)))
print(digits)
0    [3001234]
1    [1005678]
Name: col1, dtype: object

df['col1'] = digits.str.get(0).astype(int)
print(df)
col1
0  3001234
1  1005678

print(df.dtypes)
col1    int32
dtype: object

Answer 2

df2.columns [3]表示列名而不是列内容。列名没有map或apply等方法。使用df.iloc [：，column_number]或df ['column_name']来获取列的内容。

import pandas as pd
data = [u'11212; xxxxxxxxxx xxxxxxxx', 
u'11212; xxxxxxxxxx xxxxxxxx',   
u'11212; xxxxxxxxxx xxxxxxxx',   
u'11212; xxxxxxxxxx xxxxxxxx',     
u'667788; xxxxxxx xxxxxxxxxxxxx xxxxxx',   
u'55555; xxxxxxx xxxxxxxxxxxxx xxxxxx',  
u'55555; xxxxxxx xxxxxxxxxxxxx xxxxxx',   
u'55555; xxxxxxx xxxxxxxxxxxxx xxxxxx',   
u'33333; xxxxxxx xxxxxxxxxxxxx xxxxxx',   
u'333; xxx xxxxx @ xxx xxx 2 xxxx',   
u'9991; xxxx; xxxxxx xxxxx xxxx @ 2 xxx',   
u'1635; vvvvvvvvvvvv vvvvvv 10',   
u'1635; vvvvvvvvvvvv vvvvvv 10',   
u'1635; vvvvvvvvvvvv vvvvvv 10',   
u'1635; vvvvvvvvvvvv vvvvvv 10',     
u'1712; Cxxxx xxxxxxxx; xxx 0',  
u'1712; Cxxxx xxxxxxxx; xxx 0',   
u'1712; Cxxxx xxxxxxxx; xxx 0',   
u'1712; Cxxxx xxxxxxxx; xxx 0']

# make a dataframe from data as the first column
df = pd.DataFrame({'col0': data})

print df.head()

#Here I use the  iloc to the get the contents of first column (0 th column), in your case, it will 3)
df.iloc[:,0] = df.iloc[:,0].apply(lambda x: x.split(';')[0])

# in your case it will be 
#df.iloc[:,3] = df.iloc[:,3].apply(lambda x: x.split(';')[0])

print df.head()

结果

                                  col0
0            11212; xxxxxxxxxx xxxxxxxx
1            11212; xxxxxxxxxx xxxxxxxx
2            11212; xxxxxxxxxx xxxxxxxx
3            11212; xxxxxxxxxx xxxxxxxx
4  667788; xxxxxxx xxxxxxxxxxxxx xxxxxx
     col0
0   11212
1   11212
2   11212
3   11212
4  667788

Answer 3

df["Col"] = df["Col"].str.extract('(\d+)')

如何使用pandas删除列中单元格的部分内容（Python）

3 个答案: