问题描述
例如,我有一个简单的DF:
For example I have simple DF:
import pandas as pd
from random import randint
df = pd.DataFrame({'A': [randint(1, 9) for x in xrange(10)],
'B': [randint(1, 9)*10 for x in xrange(10)],
'C': [randint(1, 9)*100 for x in xrange(10)]})
我可以使用Pandas的方法和惯用法从'A'中选择与B对应的值大于50的值,对于C对应的值大于900的值吗?
Can I select values from 'A' for which corresponding values for 'B' will be greater than 50, and for 'C' - not equal 900, using methods and idioms of Pandas?
推荐答案
当然!设置:
>>> import pandas as pd
>>> from random import randint
>>> df = pd.DataFrame({'A': [randint(1, 9) for x in range(10)],
'B': [randint(1, 9)*10 for x in range(10)],
'C': [randint(1, 9)*100 for x in range(10)]})
>>> df
A B C
0 9 40 300
1 9 70 700
2 5 70 900
3 8 80 900
4 7 50 200
5 9 30 900
6 2 80 700
7 2 80 400
8 5 80 300
9 7 70 800
我们可以应用列操作并获取布尔Series对象:
We can apply column operations and get boolean Series objects:
>>> df["B"] > 50
0 False
1 True
2 True
3 True
4 False
5 False
6 True
7 True
8 True
9 True
Name: B
>>> (df["B"] > 50) & (df["C"] == 900)
0 False
1 False
2 True
3 True
4 False
5 False
6 False
7 False
8 False
9 False
[更新,以切换到新样式.loc
]:
[Update, to switch to new-style .loc
]:
然后我们可以使用它们来索引对象.对于读取访问,您可以链接索引:
And then we can use these to index into the object. For read access, you can chain indices:
>>> df["A"][(df["B"] > 50) & (df["C"] == 900)]
2 5
3 8
Name: A, dtype: int64
但是由于视图和执行写访问的副本之间的差异,您可能会遇到麻烦.您可以改用.loc
:
but you can get yourself into trouble because of the difference between a view and a copy doing this for write access. You can use .loc
instead:
>>> df.loc[(df["B"] > 50) & (df["C"] == 900), "A"]
2 5
3 8
Name: A, dtype: int64
>>> df.loc[(df["B"] > 50) & (df["C"] == 900), "A"].values
array([5, 8], dtype=int64)
>>> df.loc[(df["B"] > 50) & (df["C"] == 900), "A"] *= 1000
>>> df
A B C
0 9 40 300
1 9 70 700
2 5000 70 900
3 8000 80 900
4 7 50 200
5 9 30 900
6 2 80 700
7 2 80 400
8 5 80 300
9 7 70 800
请注意,我不小心键入了== 900
而不是!= 900
或~(df["C"] == 900)
,但是我懒得修复它.为读者练习. :^)
Note that I accidentally typed == 900
and not != 900
, or ~(df["C"] == 900)
, but I'm too lazy to fix it. Exercise for the reader. :^)
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