Question

我想在test_df中添加一个新列，其中包含取决于change_col的a或b列的值，如果change为True。下面的for循环有效，但速度太慢。 如何使用Apply 或类似方法添加新列？

test_df = pd.DataFrame({"a":[1,1,2,3],
                    "b":["ant","ber","cas","dor"],
                    "change_col":["a","b","b","a"],
                    "change":[True,True,True,False]})

    a   b      change_col   change
0   1   ant        a       True
1   1   ber        b       True
2   2   cas        b       True
3   3   dor        a       False

所需的df：

    a   b     change_col    change  new_value
0   1   ant        a        True    1
1   1   ber        b        True    ber
2   2   cas        b        True    cas
3   3   dor        a        False   NaN

我的循环

new_value= []
for _ , row in test_df.iterrows():
    if row["change"] is True:
        new_value +=[row[row["change_column"]]]
    else:
        new_value += [np.NaN]
test_df["new_value"] = new_value

我在python 3.7上使用了熊猫0.24.2。

Answer 1

您可以使用[DataFrame.lookup] [1]，

test_df['new_val'] = test_df.lookup(test_df.index, test_df['change_col'])

    a   b   change_col  change  new_val
0   1   ant a           True    1
1   1   ber b           True    ber
2   2   cas b           True    cas
3   3   dor a           False   3

编辑：要说明更改列，请使用条件

test_df['new_val'] = np.where(test_df['change'], test_df.lookup(test_df.index, test_df['change_col']), np.nan)

    a   b   change_col  change  new_val
0   1   ant a           True    1
1   1   ber b           True    ber
2   2   cas b           True    cas
3   3   dor a           False   NaN

Answer 2

由于您有多个条件，因此我们可以在此处使用np.select定义条件，并根据这些条件选择我们的值：

conditions = [
    test_df['change_col'].eq('a') & test_df['change'].eq(True),
    test_df['change_col'].eq('b') & test_df['change'].eq(True)
]

test_df['new_value'] = np.select(conditions, choicelist=[test_df['a'], test_df['b']], default=np.NaN)

输出

   a    b change_col  change new_value
0  1  ant          a    True         1
1  1  ber          b    True       ber
2  2  cas          b    True       cas
3  3  dor          a   False       NaN

Answer 3

这是使用np.select的解决方案：

import pandas as pd
import numpy as np

test_df = pd.DataFrame({"a": [1, 1, 2, 3],
                        "b": ["ant", "ber", "cas", "dor"],
                        "change_col": ["a", "b", "b", "a"],
                        "change": [True, True, True, False]})

change_a = ((test_df['change']) & (test_df['change_col'] == 'a'))
change_b = ((test_df['change']) & (test_df['change_col'] == 'b'))
dont_change = ~test_df['change']

conditions = [change_a, change_b, dont_change]
choices = [test_df['a'], test_df['b'], np.nan]

test_df["new_value"] = np.select(conditions, choices)

print(test_df)

输出：

   a    b  change change_col new_value
0  1  ant    True          a         1
1  1  ber    True          b       ber
2  2  cas    True          b       cas
3  3  dor   False          a       NaN

使用Apply根据多个列值添加新列

3 个答案: