Question

输入

      DBN Grade   3   4   5  Enrollment
0  01M015     3  30  44  15  30
1  01M015     4  30  44  15  44
2  01M015     5  30  44  15  15

所需的输出

import pandas as pd
import numpy as np

data={'DBN':['01M015','01M015','01M015'],
      'Grade':['3','4','5'],
      '3':['30','30','30'],
      '4':['44','44','44'],
      '5':['15','15','15']}

df = pd.DataFrame(data)

# This line below doesn't work: raises ValueError: Length of values does not match length of index
df['Enrollment'] = [df[c] if (df.loc[i,'Grade'] == c) else None for i in df.index for c
                    in df.columns]

您将如何创建“注册”列？

请注意，我们为每个记录寻找的列取决于df ['Grade']的值。

我尝试了df [df ['Grade']]的variations，以便可以找到df ['3']列，但是没有成功。

有没有办法做到这一点？

<html>
  <body>
    <script>
      function multa(a, b) {
        return c; 
        var a = 3;
        var b = 4; 
      }
      alert(a * b + "is" + c);
    </script>
  </body>
</html>

Answer 1

设置索引，然后使用lookup：

df.set_index('Grade').lookup(df['Grade'], df['Grade'])

array(['30', '44', '15'], dtype=object)

如果数据为数字（在示例数据中为所有字符串），则可能会遇到一些问题，需要进行强制转换才能使查询成功。

Answer 2

import pandas as pd
import numpy as np

data={'DBN':['01M015','01M015','01M015'],
      'Grade':['3','4','5'],
      '3':['30','30','30'],
      '4':['44','44','44'],
      '5':['15','15','15']}

df = pd.DataFrame(data)
enrollmentList = []

for index, row in df.iterrows():
    enrollmentList.append(row[row["Grade"]])


df['Enrollment'] =  enrollmentList

取一个单元格的值来表示熊猫的列名

2 个答案: