pandas 熊猫聚合的条件总和
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原文地址: http://stackoverflow.com/questions/15259547/
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conditional sums for pandas aggregate
提问by stites
I just recently made the switch from R to python and have been having some trouble getting used to data frames again as opposed to using R's data.table. The problem I've been having is that I'd like to take a list of strings, check for a value, then sum the count of that string- broken down by user. So I would like to take this data:
我最近才从 R 切换到 python，并且在重新习惯数据帧方面遇到了一些麻烦，而不是使用 R 的 data.table。我一直遇到的问题是我想获取一个字符串列表，检查一个值，然后对该字符串的计数求和 - 按用户细分。所以我想拿这个数据：
A_id B C
1: a1 "up" 100
2: a2 "down" 102
3: a3 "up" 100
3: a3 "up" 250
4: a4 "left" 100
5: a5 "right" 102
And return:
并返回：
A_id_grouped sum_up sum_down ... over_200_up
1: a1 1 0 ... 0
2: a2 0 1 0
3: a3 2 0 ... 1
4: a4 0 0 0
5: a5 0 0 ... 0
Before I did it with the R code (using data.table)
在我使用 R 代码（使用 data.table）之前
>DT[ ,list(A_id_grouped, sum_up = sum(B == "up"),
+ sum_down = sum(B == "down"),
+ ...,
+ over_200_up = sum(up == "up" & < 200), by=list(A)];
However all of my recent attempts with Python have failed me:
然而，我最近对 Python 的所有尝试都失败了：
DT.agg({"D": [np.sum(DT[DT["B"]=="up"]),np.sum(DT[DT["B"]=="up"])], ...
"C": np.sum(DT[(DT["B"]=="up") & (DT["C"]>200)])
})
Thank you in advance! it seems like a simple question however I couldn't find it anywhere.
先感谢您！这似乎是一个简单的问题，但我在任何地方都找不到。
采纳答案by Garrett
To complement unutbu's answer, here's an approach using applyon the groupby object.
为了补充 unutbu 的答案，这里有一种apply在 groupby 对象上使用的方法。
>>> df.groupby('A_id').apply(lambda x: pd.Series(dict(
sum_up=(x.B == 'up').sum(),
sum_down=(x.B == 'down').sum(),
over_200_up=((x.B == 'up') & (x.C > 200)).sum()
)))
over_200_up sum_down sum_up
A_id
a1 0 0 1
a2 0 1 0
a3 1 0 2
a4 0 0 0
a5 0 0 0
回答by unutbu
There might be a better way; I'm pretty new to pandas, but this works:
可能有更好的方法；我对Pandas很陌生，但这有效：
import pandas as pd
import numpy as np
df = pd.DataFrame({'A_id':'a1 a2 a3 a3 a4 a5'.split(),
'B': 'up down up up left right'.split(),
'C': [100, 102, 100, 250, 100, 102]})
df['D'] = (df['B']=='up') & (df['C'] > 200)
grouped = df.groupby(['A_id'])
def sum_up(grp):
return np.sum(grp=='up')
def sum_down(grp):
return np.sum(grp=='down')
def over_200_up(grp):
return np.sum(grp)
result = grouped.agg({'B': [sum_up, sum_down],
'D': [over_200_up]})
result.columns = [col[1] for col in result.columns]
print(result)
yields
产量
sum_up sum_down over_200_up
A_id
a1 1 0 0
a2 0 1 0
a3 2 0 1
a4 0 0 0
a5 0 0 0
