假设我有一个数据帧:
| Do | Re | Mi | Fa | So |
|---|---|---|---|---|
| 1 | 0 | Foo | 100 | 50 |
| 1 | 1 | Bar | 75 | 20 |
| 0 | 0 | True | 59 | 59 |
| 1 | 1 | False | 0 | 12 |
如何查找"Do"和"Re"列中的值都等于1且"Fa"大于"So"的所有行？
我try 了几种方法，但它们首先返回一个错误，抱怨含糊不清:
我也试着把它分解成几个步骤，但我意识到最后一步将导致我提出正确和错误的陈述.我只想要真实.
df['Do_1'] = df['Do'] == 1
df['Re_1'] = df['Re'] == 1
# This is where I realised I'm bringing in the False rows too
df['both_1'] = (df['Do1'] == df['Re_1'])```
按位AND的链掩码再增加&:
df['both_1']= (df['Fa'] > df['So']) & (df['Do'] == 1) & (df['Re'] == 1)
print (df)
Do Re Mi Fa So both_1
0 1 0 Foo 100 50 False
1 1 1 Bar 75 20 True
2 0 0 True 59 59 False
3 1 1 False 0 12 False
或者，如果可能，列表中有多个列，请按子集df[['Do', 'Re']]筛选列，并按DataFrame.all测试所有列是否为真:
df['both_1']= (df['Fa'] > df['So']) & (df[['Do', 'Re']] == 1).all(axis=1)
如果需要过滤器，请使用boolean indexing:
df1 = df[(df['Fa'] > df['So']) & (df['Do'] == 1) & (df['Re'] == 1)]
对于第二种解决方案:
df1 = df[(df['Fa'] > df['So']) & (df[['Do', 'Re']] == 1).all(axis=1)]