pandas 熊猫：组内最大值和最小值之间的差异
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Pandas: Difference between largest and smallest value within group
提问by David
Given a data frame that looks like this
给定一个看起来像这样的数据框
GROUP VALUE
1 5
2 2
1 10
2 20
1 7
I would like to compute the difference between the largest and smallest value within each group. That is, the result should be
我想计算每组中最大值和最小值之间的差异。也就是说，结果应该是
GROUP DIFF
1 5
2 18
What is an easy way to do this in Pandas?
在 Pandas 中有什么简单的方法可以做到这一点？
What is a fast way to do this in Pandas for a data frame with about 2 million rows and 1 million groups?
在 Pandas 中，对于大约有 200 万行和 100 万个组的数据框，有什么快速的方法可以做到这一点？
回答by piRSquared
Using @unutbu 's df
使用 @unutbu 的 df
per timing
unutbu's solution is best over large data sets
每个时间
unutbu 的解决方案最适合大型数据集
import pandas as pd
import numpy as np
df = pd.DataFrame({'GROUP': [1, 2, 1, 2, 1], 'VALUE': [5, 2, 10, 20, 7]})
df.groupby('GROUP')['VALUE'].agg(np.ptp)
GROUP
1 5
2 18
Name: VALUE, dtype: int64
np.ptpdocsreturns the range of an array
np.ptpdocs返回数组的范围
timing
small df
定时
小df
large dfdf = pd.DataFrame(dict(GROUP=np.arange(1000000) % 100, VALUE=np.random.rand(1000000)))
大的 dfdf = pd.DataFrame(dict(GROUP=np.arange(1000000) % 100, VALUE=np.random.rand(1000000)))
large df
many groupsdf = pd.DataFrame(dict(GROUP=np.arange(1000000) % 10000, VALUE=np.random.rand(1000000)))
大df
许多组df = pd.DataFrame(dict(GROUP=np.arange(1000000) % 10000, VALUE=np.random.rand(1000000)))
回答by unutbu
groupby/agggenerally performs best when you take advantage of the built-in aggregators such as 'max'and 'min'. So to obtain the difference, first compute the maxand minand then subtract:
groupby/agg当您利用内置的聚合如通常表现最好'max'和'min'。因此获得的区别，首先计算max和min，然后减去：
import pandas as pd
df = pd.DataFrame({'GROUP': [1, 2, 1, 2, 1], 'VALUE': [5, 2, 10, 20, 7]})
result = df.groupby('GROUP')['VALUE'].agg(['max','min'])
result['diff'] = result['max']-result['min']
print(result[['diff']])
yields
产量
diff
GROUP
1 5
2 18
回答by ASGM
You can use groupby(), min(), and max():
您可以使用groupby()，min()以及max()：
df.groupby('GROUP')['VALUE'].apply(lambda g: g.max() - g.min())
