我正在按县列出前三名的作物表。一些县以相同的顺序拥有相同的农作物品种。其他县的相同农作物品种顺序不同。
df1 = pd.DataFrame( {
"County" : ["Harney", "Baker", "Wheeler", "Hood River", "Wasco" , "Morrow","Union","Lake"] ,
"Crop1" : ["grain", "melons", "melons", "apples", "pears", "raddish","pears","pears"],
"Crop2" : ["melons","grain","grain","melons","carrots","pears","carrots","carrots"],
"Crop3": ["apples","apples","apples","grain","raddish","carrots","raddish","raddish"],
"Total_pop": [2000,1500,3000,1500,2000,2500,2700,2000]} )
我可以对Crop1，Crop2和Crop3进行分组，并获得total_pop的总和：
df1_grouped=df1.groupby(['Crop1',"Crop2","Crop3"])['Total_pop'].sum().reset_index()
这给了我特定农作物组合的总数：
df1_grouped
apples melons grain 1500
grain melons apples 2000
melons grain apples 4500
pears carrots raddish 6700
raddish pears carrots 2500
不过，我希望获得的是不同作物组合的总人口-不论列出的作物是crop1，crop2还是crop3。期望的结果将是这样：
apples melons grain 8000
pears carrots raddish 9200
谢谢您的指导。
答案 0 :(得分：4)
由于您的数据似乎可以保证每个国家/地区拥有3种独特的农作物（“我正在按县级列出前三名农作物的表格。”），因此可以对这些值进行排序并重新分配。
import numpy as np
cols = ['Crop1', 'Crop2', 'Crop3']
df1[cols] = np.sort(df1[cols].values, axis=1)
County Crop1 Crop2 Crop3 Total_pop
0 Harney apples grain melons 2000
1 Baker apples grain melons 1500
2 Wheeler apples grain melons 3000
3 Hood River apples grain melons 1500
4 Wasco carrots pears raddish 2000
5 Morrow carrots pears raddish 2500
6 Union carrots pears raddish 2700
7 Lake carrots pears raddish 2000
然后总结：
df1.groupby(cols).sum()
# Total_pop
#Crop1 Crop2 Crop3
#apples grain melons 8000
#carrots pears raddish 9200
好处是您避免使用Series.apply或.apply(axis=1)。对于较大的DataFrames，性能差异非常明显：
df1 = pd.concat([df1]*10000, ignore_index=True)
cols = ['Crop1', 'Crop2', 'Crop3']
%timeit df1[cols] = np.sort(df1[cols].values, axis=1)
#36.1 ms ± 399 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
to_sum = ['Crop1', 'Crop2', 'Crop3']
%timeit df1[to_sum] = pd.DataFrame(df1.loc[:, to_sum].apply(set, axis=1).apply(list).values.tolist(), columns=to_sum)
#1.41 s ± 51.2 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
答案 1 :(得分：2)
这是一种方法。
首先让我们跨列获取唯一值，然后将这些值重新分配给DataFrame。我们将在原始数据的副本上执行此操作，因为您可能需要保留原始数据。
df = df1.copy()
to_sum = ['Crop1', 'Crop2', 'Crop3']
df[to_sum] = pd.DataFrame(df.loc[:, to_sum] \
.apply(set, axis=1) \
.apply(sorted) \
.values \
.tolist(), columns=to_sum)
print(df)
County Crop1 Crop2 Crop3 Total_pop
0 Harney grain apples melons 2000
1 Baker grain apples melons 1500
2 Wheeler grain apples melons 3000
3 Hood River grain apples melons 1500
4 Wasco pears carrots raddish 2000
5 Morrow pears carrots raddish 2500
6 Union pears carrots raddish 2700
7 Lake pears carrots raddish 2000
现在我们可以执行groupby以获得所需的结果。
df.groupby(to_sum).Total_pop.sum()
Crop1 Crop2 Crop3
apples grain melons 8000
carrots pears raddish 9200
Name: Total_pop, dtype: int64
答案 2 :(得分：2)
np.bincount i, u = pd.factorize([*map(frozenset, zip(df1.Crop1, df1.Crop2, df1.Crop3))])
s = np.bincount(i, df1.Total_pop)
pd.Series(s, u)
(melons, grain, apples) 8000.0
(carrots, raddish, pears) 9200.0
dtype: float64
或者，如果您想要单独的列
pd.Series(dict(zip(map(tuple, u), s)))
melons grain apples 8000.0
carrots raddish pears 9200.0
dtype: float64
又漂亮
pd.Series(dict(zip(map(tuple, u), s))) \
.rename_axis(['Crop1', 'Crop2', 'Crop3']).reset_index(name='Total_pop')
Crop1 Crop2 Crop3 Total_pop
0 melons grain apples 8000.0
1 carrots raddish pears 9200.0
答案 3 :(得分：1)
方法1：
合并crop列
>>> df1['combined_temp'] = df1.apply(lambda x : list([x['Crop1'],
... x['Crop2'],
... x['Crop3']]),axis=1)
>>> df1.head()
County Crop1 Crop2 Crop3 Total_pop combined_temp
0 Harney grain melons apples 2000 [grain, melons, apples]
1 Baker melons grain apples 1500 [melons, grain, apples]
2 Wheeler melons grain apples 3000 [melons, grain, apples]
3 Hood River apples melons grain 1500 [apples, melons, grain]
4 Wasco pears carrots raddish 2000 [pears, carrots, raddish]
将其设为已排序的元组
>>> df1['sorted'] = df1.apply(lambda x : tuple(sorted(x['combined_temp'])),axis=1)
>>> df1.head()
County Crop1 Crop2 ... Total_pop combined_temp sorted
0 Harney grain melons ... 2000 [grain, melons, apples] (apples, grain, melons)
1 Baker melons grain ... 1500 [melons, grain, apples] (apples, grain, melons)
2 Wheeler melons grain ... 3000 [melons, grain, apples] (apples, grain, melons)
3 Hood River apples melons ... 1500 [apples, melons, grain] (apples, grain, melons)
4 Wasco pears carrots ... 2000 [pears, carrots, raddish] (carrots, pears, raddish)
然后通过操作进入正常的小组
>>> df1_grouped = df1.groupby(['sorted'])['Total_pop'].sum().reset_index()
>>> df1_grouped
sorted Total_pop
0 (apples, grain, melons) 8000
1 (carrots, pears, raddish) 9200
方法2： 由answer到aws-apprentice的缩写版本
df = df1.copy()
grouping_cols = ['Crop1', 'Crop2', 'Crop3']
df[grouping_cols] = pd.DataFrame(df.loc[:, grouping_cols] \
.apply(set, axis=1) \
.apply(sorted)
.values \
.tolist(), columns=grouping_cols)
>>> df.head()
County Crop1 Crop2 Crop3 Total_pop
0 Harney apples grain melons 2000
1 Baker apples grain melons 1500
2 Wheeler apples grain melons 3000
3 Hood River apples grain melons 1500
4 Wasco carrots pears raddish 2000
现在按组分组
>>> df.groupby(grouping_cols).Total_pop.sum()
Crop1 Crop2 Crop3
apples grain melons 8000
carrots pears raddish 9200
Name: Total_pop, dtype: int64
但我个人更喜欢this answer using numpy
答案 4 :(得分：0)
import pandas as pd
df = pd.DataFrame( {
"County" : ["Harney", "Baker", "Wheeler", "Hood River", "Wasco" , "Morrow","Union","Lake"] ,
"Crop1" : ["grain", "melons", "melons", "apples", "pears", "raddish","pears","pears"],
"Crop2" : ["melons","grain","grain","melons","carrots","pears","carrots","carrots"],
"Crop3": ["apples","apples","apples","grain","raddish","carrots","raddish","raddish"],
"Total_pop": [2000,1500,3000,1500,2000,2500,2700,2000]} )
print(df)
df["Merged"] = df[["Crop1", "Crop2", "Crop3"]].apply(lambda x: ','.join(x.dropna().astype(str).values), axis=1).str.split(",")
df["Merged"] = df["Merged"].sort_values().apply(lambda x: sorted(x)).apply(lambda x: ",".join(x))
df[["x", "y", "z"]] = df["Merged"].str.split(",", expand=True)
df1=df.groupby(['x',"y","z"])['Total_pop'].sum().reset_index()
print(df1)
输出：
County Crop1 Crop2 Crop3 Total_pop
Harney grain melons apples 2000
Baker melons grain apples 1500
Wheeler melons grain apples 3000
Hood River apples melons grain 1500
Wasco pears carrots raddish 2000
Morrow raddish pears carrots 2500
Union pears carrots raddish 2700
Lake pears carrots raddish 2000
x y z Total_pop
apples grain melons 8000
carrots pears raddish 9200