Python 计算在所有行中使用lambda函数填充值（NULL除外）的字段数

谢谢你看这个问题。我正在使用lambda创建一个逻辑，该逻辑贯穿所有行，并统计除NA之外具有值的字段数。正如您在给定的示例中所看到的

Input : 

project_id   project_a   project_b   project_c   project_d   project_e
     1             Yes         Yes        Yes         NA       Yes
     2             Yes         Yes        Yes         NA       Yes
     3             NA          Yes        Yes         NA       Yes
     4             Yes         Yes        Yes         NA       Yes
     5             NA          Yes        Yes         NA       Yes

Desired Output :

project_id   project_a   project_b   project_c   project_d   project_e    field_populated
     1             Yes         Yes        Yes         NA       Yes              5
     2             Yes         Yes        Yes         NA       Yes              5
     3             NA          Yes        Yes         NA       Yes              3
     4             Yes         Yes        Yes         NA       Yes              5
     5             NA          Yes        Yes         NA       Yes              4

我尝试过使用以下代码，但遇到了一些问题

proj_table['field_populated'] = proj_table['project_id', 'project_a', 'project_b','project_c, 'project_d','project_e].apply(lambda x: x+1 if x != "NA" or np.nan else x) 

---

如果将其过度复杂化，则可以使用、

count

计算数据帧的非空值并填充新列，并沿行执行操作（

axis=1

）

<代码>过滤器（like＝Project） >只考虑“Project”列，以防您的实际<代码> df中有更多列。

df['field_populated'] = df.filter(like='project').count(axis=1)

其中打印：

df

   project_id project_a project_b  ... project_d  project_e field_populated
0           1       Yes       Yes  ...       NaN        Yes               5
1           2       Yes       Yes  ...       NaN        Yes               5
2           3       NaN       Yes  ...       NaN        Yes               4
3           4       Yes       Yes  ...       NaN        Yes               5
4           5       NaN       Yes  ...       NaN        Yes               4

  col1 col2 col3 col4 col5  field_populated
0    1    2    3    4    5                5
1    1    2   NA    4    5                4
2    1    2    3    4   NA                4
3   NA    2   NA    4    5                3

---

正如sophocles所说，这里不需要lambda函数。但是，如果您出于学习的原因打算应用lambda函数，您可以执行一个简单的lambda，计算每行中有多少个非“X”值。此外，如果您想做得更“lambdish”，您也可以用另一个函数检查值是否为“NA”

这里有一个小例子：

import pandas as pd

not_NA = lambda column: 1 if column != "NA" else 0

count_NA = lambda row: sum(row.apply(not_NA))

df = pd.DataFrame([["1","2","3","4","5"], ["1","2","NA","4","5"], ["1","2","3","4","NA"], ["NA","2","NA","4","5"]], columns=list(["col1","col2","col3","col4","col5"]))

df['field_populated'] = df.apply(count_NA, axis=1)

其中打印：

df

   project_id project_a project_b  ... project_d  project_e field_populated
0           1       Yes       Yes  ...       NaN        Yes               5
1           2       Yes       Yes  ...       NaN        Yes               5
2           3       NaN       Yes  ...       NaN        Yes               4
3           4       Yes       Yes  ...       NaN        Yes               5
4           5       NaN       Yes  ...       NaN        Yes               4

  col1 col2 col3 col4 col5  field_populated
0    1    2    3    4    5                5
1    1    2   NA    4    5                4
2    1    2    3    4   NA                4
3   NA    2   NA    4    5                3

---

请注意，在lambda函数“not_NA”中，您还可以添加其他值，以便在需要时丢弃。

我猜您输入了一个错误，

project_id=3

的值应该是

4

？是的，有@ShubhamSharma。谢谢