假设我有这个数据框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column2中每一行的所有字符的总和(不包括空格),然后将其返回到新的column4中,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中每一行所有字符的总和:
df['column4'] = sum(list(map(lambda x : sum(len(y) for y in x.split()), df['column2'])))
所以目前我的df看起来像这样:
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
有人有主意吗?
参考方案
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()))
或获取所有值的计数并通过Series.str.count
减去空格计数:
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
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