pandas 熊猫数据框返回列字符串中的第一个单词
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pandas dataframe return first word in string for column
提问by Testy8
I have a dataframe:
我有一个数据框：
df = pd.DataFrame({'id' : ['abarth 1.4 a','abarth 1 a','land rover 1.3 r','land rover 2',
'land rover 5 g','mazda 4.55 bl'],
'series': ['a','a','r','','g', 'bl'] })
I would like to remove the 'series' string from the corresponding id, so the end result should be:
我想从相应的 id 中删除“系列”字符串，因此最终结果应该是：
Final result should be 'id': ['abarth 1.4','abarth 1','land rover 1.3','land rover 2','land rover 5', 'mazda 4.55']
最终结果应该是 'id': ['abarth 1.4','abarth 1','land rover 1.3','land rover 2','land rover 5', 'mazda 4.55']
Currently I am using df.apply:
目前我正在使用 df.apply：
df.id = df.apply(lambda x: x['id'].replace(x['series'], ''), axis =1)
But this removes all instances of the strings, even in other words, like so:
'id': ['brth 1.4','brth 1','land ove 1.3','land rover 2','land rover 5', 'mazda 4.55']
但这会删除字符串的所有实例，换句话说，就像这样：
'id': ['brth 1.4','brth 1','land ove 1.3','land rover 2','land rover 5', 'mazda 4.55']
Should I somehow mix and match regex with the variable inside df.apply, like so?
我应该以某种方式将正则表达式与 df.apply 中的变量混合和匹配吗？
df.id = df.apply(lambda x: x['id'].replace(r'\b' + x['series'], ''), axis =1)
回答by piRSquared
Use str.splitand str.getand assign using loconly where df.make == ''
使用str.split和str.get分配loc仅使用wheredf.make == ''
df.loc[df.make == '', 'make'] = df.id.str.split().str.get(0)
print df
id make
0 abarth 1.4 abarth
1 abarth 1 abarth
2 land rover 1.3 rover
3 land rover 2 rover
4 land rover 5 rover
5 mazda 4.55 mazda
回答by Parfait
Consider a regex solution with locwhere it extracts everything before first space:
考虑一个正则表达式解决方案loc，它在第一个空间之前提取所有内容：
df.loc[df['make']=='', 'make'] = df['id'].str.extract('(.*) ', expand=False)
Alternatively, use numpy's wherewhich allows the if/then/else conditional logic:
或者，使用 numpy'swhere允许 if/then/else 条件逻辑：
df['make'] = np.where(df['make']=='',
df['id'].str.extract('(.*) ', expand=False),
df['make'])
回答by Aamir Khan
If I got your question correctly you can just use replacefunction:
如果我正确回答了您的问题，您可以使用replace函数：
df.make = df.make.replace("", test.id)
回答by Qazi Basheer
It's simple. Use as follows:
这很简单。使用方法如下：
df['make'] = df['id'].str.split(' ').str[0]
