Python 基于另一列有条件地更改数据帧列'；s值

我有以下数据帧

d1

：

+----------+-------+---------+--------------+
| Item Num | Cost  | Revenue |  Rev / Cost  |
+----------+-------+---------+--------------+
|        1 | 45.76 |  345.67 | 7.5539772727 |
|        2 | 55.78 |  456.92 | 8.1914664754 |
|        3 | 34.68 |       0 |            0 |
|        4 | 79.85 |       0 |            0 |
+----------+-------+---------+--------------+

我希望

Cost/Rev

列的值等于该行的

Cost

，如果“Cost/Rev”等于0，则乘以负1

因此，期望的输出是：

+----------+-------+---------+--------------+
| Item Num | Cost  | Revenue |  Rev / Cost  |
+----------+-------+---------+--------------+
|        1 | 45.76 |  345.67 | 7.5539772727 |
|        2 | 55.78 |  456.92 | 8.1914664754 |
|        3 | 34.68 |       0 |       -34.68 |
|        4 | 79.85 |       0 |       -79.85 |
+----------+-------+---------+--------------+

到目前为止，我得到的是：

d1['Rev / Cost'] = d1['Rev / Cost'].apply(lambda x: x if x > 0 else d1['Cost'])

它只是用一个值覆盖预期范围，并抛出以下警告：

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

---

创建一个掩码，然后使用

loc

分配给子掩码

mask = df['Rev / Cost'] == 0
df.loc[mask, 'Rev / Cost'] = df.loc[mask, 'Cost'].mul(-1)

---

创建一个掩码，然后使用

loc

分配给子掩码

mask = df['Rev / Cost'] == 0
df.loc[mask, 'Rev / Cost'] = df.loc[mask, 'Cost'].mul(-1)

---

由于布尔值计算为0/1，您只需将条件乘以成本，然后将其从Rev/Cost中减去即可。这将大大提高性能

df['Rev / Cost'] -=  df['Cost'] * (df['Rev / Cost'] == 0)

您也可以使用

np.where

df['Rev / Cost'] = np.where(df['Rev / Cost'] == 0, -df['Cost'], df['Rev / Cost']

df['Rev / Cost'] = df['Rev / Cost'].where(lambda x: x != 0, df.Cost)

或

系列。其中

df['Rev / Cost'] = np.where(df['Rev / Cost'] == 0, -df['Cost'], df['Rev / Cost']

df['Rev / Cost'] = df['Rev / Cost'].where(lambda x: x != 0, df.Cost)

---

由于布尔值计算为0/1，您只需将条件乘以成本，然后将其从Rev/Cost中减去即可。这将大大提高性能

df['Rev / Cost'] -=  df['Cost'] * (df['Rev / Cost'] == 0)

您也可以使用

np.where

df['Rev / Cost'] = np.where(df['Rev / Cost'] == 0, -df['Cost'], df['Rev / Cost']

df['Rev / Cost'] = df['Rev / Cost'].where(lambda x: x != 0, df.Cost)

或

系列。其中

df['Rev / Cost'] = np.where(df['Rev / Cost'] == 0, -df['Cost'], df['Rev / Cost']

df['Rev / Cost'] = df['Rev / Cost'].where(lambda x: x != 0, df.Cost)