![](/img/trans.png)
[英]Create a new pandas dataframe column based on other column of the dataframe
[英]How to create new column based on substrings in other column in a pandas dataframe?
我有一个 dataframe 结构如下:
df = pd.DataFrame({
'Substance': ['(NPK) 20/10/6', '(NPK) Guayacan 10/20/30', '46%N / O%P2O5 (Urea)', '46%N / O%P2O5 (Urea)', '(NPK) DAP Diammonphosphat; 18/46/0'],
'value': [0.2, 0.4, 0.6, 0.8, .9]
})
substance value
0 (NPK) 20/10/6 0.2
1 (NPK) Guayacan 10/20/30 0.4
2 46%N / O%P2O5 (Urea) 0.6
3 46%N / O%P2O5 (Urea) 0.8
4 (NPK) DAP Diammonphosphat; 18/46/0 0.9
现在我想用物质的简称创建一个新列:
test['Short Name'] = test['Substance'].apply(lambda x: 'Urea' if
any(i in x for i in 'Urea') else '(NPK)')
最后一行代码有两个问题。 首先,output 看起来是这样的:
Substance value Short Name
0 (NPK) 20/10/6 0.2 (NPK)
1 (NPK) Guayacan 10/20/30 0.4 Urea
2 46%N / O%P2O5 (Urea) 0.6 Urea
3 46%N / O%P2O5 (Urea) 0.8 Urea
4 (NPK) DAP Diammonphosphat; 18/46/0 0.9 (NPK)
所以第二个条目也标有尿素,尽管它应该是 NPK。
此外,我的实际数据也会产生以下错误,尽管使用了原始物质名称,但有趣/令人讨厌的是,我无法用虚拟数据重现该错误。
/var/folders/tf/hzv31v4x42q4_mnw4n8ldhsm0000gn/T/ipykernel_10743/136042259.py:5: SettingWithCopyWarning:
A value is trying to be set on a copy of a slice from a DataFrame.
Try using .loc[row_indexer,col_indexer] = value instead
注意:由于我还有更多的内容,我将不得不在 if/else 循环中添加更多语句。
编辑:物质名称需要映射到以下短名称列表:
样本数据的预期 output 将是
Substance value Short Name
0 (NPK) 20/10/6 0.2 (NPK)
1 (NPK) Guayacan 10/20/30 0.4 (NPK)
2 46%N / O%P2O5 (Urea) 0.6 Urea
3 46%N / O%P2O5 (Urea) 0.8 Urea
4 (NPK) DAP Diammonphosphat; 18/46/0 0.9 (NPK)
Edit2:然后我想添加一个语句,以便我收到以下 output:
Substance value Short Name
0 (NPK) 20/10/6 0.2 (NPK)
1 (NPK) Guayacan 10/20/30 0.4 (NPK)
2 46%N / O%P2O5 (Urea) 0.6 Urea
3 46%N / O%P2O5 (Urea) 0.8 Urea
4 (NPK) DAP Diammonphosphat; 18/46/0 0.9 DAP
尝试这个:
df['Short Name'] = df['Substance'].str.extract(r'\((.+?)\)')
Output:
>>> df
Substance value Short Name
0 (NPK) 20/10/6 0.2 NPK
1 (NPK) Guayacan 10/20/30 0.4 NPK
2 46%N / O%P2O5 (Urea) 0.6 Urea
3 46%N / O%P2O5 (Urea) 0.8 Urea
4 (NPK) 20/10/6 0.9 NPK
为我工作:
df['Short Name'] = df['Substance'].apply(lambda x: 'Urea' if 'Urea' in x else '(NPK)')
>>> df
Substance value Short Name
0 (NPK) 20/10/6 0.2 (NPK)
1 (NPK) Guayacan 10/20/30 0.4 (NPK)
2 46%N / O%P2O5 (Urea) 0.6 Urea
3 46%N / O%P2O5 (Urea) 0.8 Urea
4 (NPK) 20/10/6 0.9 (NPK)
正则表达式:
import re
short = re.compile(r"\W*(urea)\W*", re.I)
df['Short Name'] = df['Substance'].apply(lambda x: 'Urea' if len(short.findall(x.lower())) else '(NPK)')
不是最整洁的解决方案,但至少是一个解决方案:
test['Short Name'] = test['Substance'].apply(lambda x: 'Urea' if 'Urea' in x else 'DAP' if 'DAP' in x else '(NPK)')
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.