本文介绍了将pandas DataFrame列扩展为多行的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
如果我有这样的DataFrame
:
pd.DataFrame( {"name" : "John",
"days" : [[1, 3, 5, 7]]
})
具有以下结构:
days name
0 [1, 3, 5, 7] John
如何将其扩展到以下内容?
How do expand it to the following?
days name
0 1 John
1 3 John
2 5 John
3 7 John
推荐答案
您可以使用df.itertuples
遍历每一行,并使用列表推导将数据重塑为所需的形式:
You could use df.itertuples
to iterate through each row, and use a list comprehension to reshape the data into the desired form:
import pandas as pd
df = pd.DataFrame( {"name" : ["John", "Eric"],
"days" : [[1, 3, 5, 7], [2,4]]})
result = pd.DataFrame([(d, tup.name) for tup in df.itertuples() for d in tup.days])
print(result)
收益
0 1
0 1 John
1 3 John
2 5 John
3 7 John
4 2 Eric
5 4 Eric
Divakar的解决方案 using_repeat
是最快的:
Divakar's solution, using_repeat
, is fastest:
In [48]: %timeit using_repeat(df)
1000 loops, best of 3: 834 µs per loop
In [5]: %timeit using_itertuples(df)
100 loops, best of 3: 3.43 ms per loop
In [7]: %timeit using_apply(df)
1 loop, best of 3: 379 ms per loop
In [8]: %timeit using_append(df)
1 loop, best of 3: 3.59 s per loop
以下是用于上述基准测试的设置:
Here is the setup used for the above benchmark:
import numpy as np
import pandas as pd
N = 10**3
df = pd.DataFrame( {"name" : np.random.choice(list('ABCD'), size=N),
"days" : [np.random.randint(10, size=np.random.randint(5))
for i in range(N)]})
def using_itertuples(df):
return pd.DataFrame([(d, tup.name) for tup in df.itertuples() for d in tup.days])
def using_repeat(df):
lens = [len(item) for item in df['days']]
return pd.DataFrame( {"name" : np.repeat(df['name'].values,lens),
"days" : np.concatenate(df['days'].values)})
def using_apply(df):
return (df.apply(lambda x: pd.Series(x.days), axis=1)
.stack()
.reset_index(level=1, drop=1)
.to_frame('day')
.join(df['name']))
def using_append(df):
df2 = pd.DataFrame(columns = df.columns)
for i,r in df.iterrows():
for e in r.days:
new_r = r.copy()
new_r.days = e
df2 = df2.append(new_r)
return df2
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