将行中所有字符的总和返回到另一列

Question

假设我有这个数据帧df：

column1      column2                                            column3
amsterdam    school yeah right backtic escapes sport swimming   2016
rotterdam    nope yeah                                          2012
thehague     i now i can fly no you cannot swimming rope        2010
amsterdam    sport cycling in the winter makes me               2019

如何获取第2列中每行所有字符（不包括空格）的总和，并将其返回到新的第4列，如下所示：

column1      column2                                            column3    column4
amsterdam    school yeah right backtic escapes sport swimming   2016       70
rotterdam    nope yeah                                          2012       8
thehague     i now i can fly no you cannot swimming rope        2010       65
amsterdam    sport cycling in the winter makes me               2019       55

我尝试了这段代码，但到目前为止，我得到了column2中每行所有字符的总和：

df['column4'] = sum(list(map(lambda x : sum(len(y) for y in x.split()), df['column2'])))

因此，当前我的df如下所示：

column1      column2                                            column3    column4
amsterdam    school yeah right backtic escapes sport swimming   2016          250
rotterdam    nope yeah                                          2012           250
thehague     i now i can fly no you cannot swimming rope        2010           250
amsterdam    sport cycling in the winter makes me               2019           250

有人知道吗

Answer 1

可以将方法count与正则表达式模式一起使用：

df['column2'].str.count(pat='\w')

输出：

0    42
1     8
2    34
3    30
Name: column2, dtype: int64

​

Answer 2

嗨，这对我有用

import pandas as pd
df=pd.DataFrame({'col1':['Stack Overflow','The Guy']})
df['Count Of Chars']=df['col1'].str.replace(" ","").apply(len)
df

输出

    col1    Count Of characters
0   Stack Overflow  13
1   The Guy          6

Answer 3

在解决方案中使用自定义lambda函数：

df['column4'] = df['column2'].apply(lambda x: sum(len(y) for y in x.split()))

或者获取所有值的计数并用^{}减去空白的计数：

df['column4'] = df['column2'].str.len().sub(df['column2'].str.count(' '))
#rewritten to custom functon
#df['column4'] = df['column2'].map(lambda x: len(x) - x.count(' '))
print (df)
     column1                                           column2  column3  \
0  amsterdam  school yeah right backtic escapes sport swimming     2016   
1  rotterdam                                         nope yeah     2012   
2   thehague       i now i can fly no you cannot swimming rope     2010   
3  amsterdam              sport cycling in the winter makes me     2019   

   column4  
0       42  
1        8  
2       34  
3       30