Python 与列表列中的项目关联的金额总和
我有一个数据框,如下所示:Python 与列表列中的项目关联的金额总和,python,pandas,dataframe,Python,Pandas,Dataframe,我有一个数据框,如下所示: country letter keywords amount c y ['fruits', 'apples', "banana"] 700 c y ["music", "dance", "banana"] 150 c y ['loud', "dance", "apples"]
country letter keywords amount
c y ['fruits', 'apples', "banana"] 700
c y ["music", "dance", "banana"] 150
c y ['loud', "dance", "apples"] 350
我想计算与每个关键字相关联的金额。
注:国家
和字母
并不总是相同的,如上面的人为数据所示。此外,关键字列表的大小也各不相同
我试过几种解决办法。我在下面附上了我最快的一个。我还尝试了apply
和defaultdicts
的解决方案
keywords_list = []
for i in zip(*[df[c] for c in df.columns]):
data = list(i[0:2])
for k in i[2]:
row = [k] + data + [i[-1]]
keywords_list.append(row)
df_expanded = pd.DataFrame(keywords_list)
df_expanded.groupby(list(range(3)))[3].sum().reset_index()
目标
country letter keywords amount
0 c y apples 1050
1 c y banana 850
2 c y dance 500
3 c y fruits 700
4 c y loud 350
5 c y music 150
编辑:纠正目标示例中的错误
country letter keywords amount
0 c y apples 1050
1 c y banana 850
2 c y dance 500
3 c y fruits 700
4 c y loud 350
5 c y music 150
数据
country = list("ccc")
letters = list("yyy")
keywords = [['fruits', 'apples', "banana"], ["music", "dance", "banana"], ['loud', "dance", "apples"]]
amount = [700, 150, 350]
df = pd.DataFrame({"country" : country, "keywords": keywords, "letter" : letters, "amount" : amount})
df = df[['country', 'letter', 'keywords', 'amount']]
试试这个:
In [76]: (df.set_index(['country','letter','amount'])
...: .keywords
...: .apply(pd.Series)
...: .stack()
...: .reset_index(name='keywords')
...: .drop('level_3',1)
...: )
...:
Out[76]:
country letter amount keywords
0 c y 700 fruits
1 c y 700 apples
2 c y 700 banana
3 c y 150 music
4 c y 150 dance
5 c y 150 banana
6 c y 350 loud
7 c y 350 dance
8 c y 350 apples
您可以使用:
df1 = pd.DataFrame(df.keywords.values.tolist())
.stack()
.reset_index(level=1, drop=True)
.rename('keywords')
print (df1)
0 fruits
0 apples
0 banana
1 music
1 dance
1 banana
2 loud
2 dance
2 apples
Name: keywords, dtype: object
print (df.drop('keywords', axis=1).join(df1).reset_index(drop=True))
country letter amount keywords
0 c y 700 fruits
1 c y 700 apples
2 c y 700 banana
3 c y 150 music
4 c y 150 dance
5 c y 150 banana
6 c y 350 loud
7 c y 350 dance
8 c y 350 apples
另一个解决方案:
df = df.set_index(['country','letter','amount'])
df1 = pd.DataFrame(df.keywords.values.tolist(), index = df.index) \
.stack() \
.reset_index(name='keywords') \
.drop('level_3',axis=1)
print (df1)
country letter amount keywords
0 c y 700 fruits
1 c y 700 apples
2 c y 700 banana
3 c y 150 music
4 c y 150 dance
5 c y 150 banana
6 c y 350 loud
7 c y 350 dance
8 c y 350 apples
然后需要使用聚合总和的groupby
:
print (df.groupby(['country','letter','keywords'], as_index=False)['amount'].sum())
country letter keywords amount
0 c y apples 1050
1 c y banana 850
2 c y dance 500
3 c y fruits 700
4 c y loud 350
5 c y music 150
计时:
In [47]: %timeit (df.set_index(['country','letter','amount']).keywords.apply(pd.Series).stack().reset_index().drop('level_3',1))
1 loop, best of 3: 4.55 s per loop
In [48]: %timeit (jez1(df3))
10 loops, best of 3: 24.8 ms per loop
In [49]: %timeit (jez2(df3))
10 loops, best of 3: 29.7 ms per loop
计时代码:
df = pd.concat([df]*10000).reset_index(drop=True)
df3 = df.copy()
df4 = df.copy()
def jez1(df):
df1 = pd.DataFrame(df.keywords.values.tolist()).stack().reset_index(level=1, drop=True).rename('keywords')
return df.drop('keywords', axis=1).join(df1).reset_index(drop=True)
def jez2(df):
df = df.set_index(['country','letter','amount'])
df1 = pd.DataFrame(df.keywords.values.tolist(), index = df.index).stack().reset_index(name='keywords').drop('level_3',axis=1)
return df1
感谢您的改进,这样就不需要再次删除。不幸的是,计时
失败(KeyError:'keywords'
),因此我无法对其进行比较 是的,这个解决方案好得多。但我认为我们可以进一步改进它:df.join(pd.DataFrame(df.pop('keywords').values.tolist()).stack().reset_index(level=1,drop=True)。rename('keywords'))
@MaxU-谢谢您的改进pop
很少使用,但在这里它是个好主意;)