英文:
Pandas filtering one column but only if another column is less than a specified value
问题
我有一个类似这样的数据框：
| C1 | C2 |
|---|---|
| 100 | 10 |
| 25 | 8 |
| 67 | 4 |
| 0 | 4 |
| 0 | 1 |
| 67 | 0 |
我尝试应用一个筛选条件，如果C2中的值大于2，则移除C1中的0值。目前我的筛选条件给出了如下输出：
| C1 | C2 |
|---|---|
| 100 | 10 |
| 25 | 8 |
| 67 | 4 |
而我想要的输出应该是：
| C1 | C2 |
|---|---|
| 100 | 10 |
| 25 | 8 |
| 67 | 4 |
| 0 | 1 |
| 67 | 0 |
我的当前筛选条件是：
(df['C1'] != 0) & (df['C2'] > 2)
但我也尝试过：
(df['C1'] > 0) & (df['C2'] > 2)
结果是相同的。
英文:
I have a dataframe that looks like this:
| C1 | C2 |
|---|---|
| 100 | 10 |
| 25 | 8 |
| 67 | 4 |
| 0 | 4 |
| 0 | 1 |
| 67 | 0 |
And I'm trying to apply a filter that removes 0 values in C1 but only if the values in C2 are greater than two. At the moment my filter gives me an output that looks like this:
| C1 | C2 |
|---|---|
| 100 | 10 |
| 25 | 8 |
| 67 | 4 |
whilst the output I want would be:
| C1 | C2 |
|---|---|
| 100 | 10 |
| 25 | 8 |
| 67 | 4 |
| 0 | 1 |
| 67 | 0 |
my filter currently is:
(df['C1'] != 0) & (df['C2'] > 2)
but I've also tried
(df['C1'] > 0) & (df['C2'] > 2)
which gave me the same result
答案1
得分: 2
使用两个掩码进行布尔索引：
out = df[~(df['C1'].eq(0) & df['C2'].gt(2))]
或者：
out = df[df['C1'].ne(0) | df['C2'].le(2)]
输出：
C1 C2
0 100 10
1 25 8
2 67 4
4 0 1
5 67 0
可复现的输入：
df = pd.DataFrame({'C1': [100, 25, 67, 0, 0, 67],
'C2': [10, 8, 4, 4, 1, 0]})
英文:
Use boolean indexing with two masks:
out = df[~(df['C1'].eq(0) & df['C2'].gt(2))]
Or:
out = df[df['C1'].ne(0) | df['C2'].le(2)]
Output:
C1 C2
0 100 10
1 25 8
2 67 4
4 0 1
5 67 0
Reproducible input:
df = pd.DataFrame({'C1': [100, 25, 67, 0, 0, 67],
'C2': [10, 8, 4, 4, 1, 0]})
答案2
得分: 2
这应该有效。
new_df = df.loc[~((df.C1 == 0) & (df.C2 > 2))]
这也应该有效。
new_df = df.loc[(df.C1 != 0) | (df.C2 <= 2)]
英文:
This should work.
new_df = df.loc[~((df.C1 == 0) & (df.C2 > 2))]
as should this one.
new_df = df.loc[(df.C1 =! 0) | (df.C2 <= 2))]
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论