Pandas: groupby 然后根据条件计数给出 NaN

Question

我有以下数据集：

+----+------+
| ID | Type |
+----+------+
| a  | New  |
+----+------+
| b  | Old  |
+----+------+
| b  | Old  |
+----+------+
| b  | New  |
+----+------+
| c  | Old  |
+----+------+

我正在尝试按 ID 分组，然后计算每个组的 New 出现次数。例如，我会有 a=1、b=2 和 c=0.

这是我尝试过的方法：

df['NewAmount'] = df.groupby('ID')['Type'].apply(
    lambda x: x[x == 'New'].count())

我明白了：

+----+------+----------+
| ID | Type | NewAmount|
+----+------+----------+
| a  | New  | NaN      |
+----+------+----------+
| b  | Old  | NaN      |
+----+------+----------+
| b  | Old  | NaN      |
+----+------+----------+
| b  | New  | NaN      |
+----+------+----------+
| c  | Old  | NaN      |
+----+------+----------+

Answer 1

你应该试试 transform

df['out'] = df['Type'].eq('New').groupby(df['ID']).transform('sum')

Pandas: groupby 然后根据条件计数给出 NaN

Pandas: groupby then count based on condition gives NaN

python

data-analysis

pandas