如何将带有值列表的列转换为 Pandas DataFrame 中的行
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How to convert column with list of values into rows in Pandas DataFrame
提问by Agus Sanjaya
Hi I have a dataframe like this:
嗨，我有一个这样的数据框：
A B
0: some value [[L1, L2]]
I want to change it into:
我想把它改成：
A B
0: some value L1
1: some value L2
How can I do that?
我怎样才能做到这一点？
采纳答案by MaxU
you can do it this way:
你可以这样做：
In [84]: df
Out[84]:
A B
0 some value [[L1, L2]]
1 another value [[L3, L4, L5]]
In [85]: (df['B'].apply(lambda x: pd.Series(x[0]))
....: .stack()
....: .reset_index(level=1, drop=True)
....: .to_frame('B')
....: .join(df[['A']], how='left')
....: )
Out[85]:
B A
0 L1 some value
0 L2 some value
1 L3 another value
1 L4 another value
1 L5 another value
UPDATE:a more generic solution
更新：更通用的解决方案
回答by Pygirl
Pandas >= 0.25
Pandas >= 0.25
df1 = pd.DataFrame({'A':['a','b'],
'B':[[['1', '2']],[['3', '4', '5']]]})
print(df1)
A B
0 a [[1, 2]]
1 b [[3, 4, 5]]
df1 = df1.explode('B')
df1.explode('B')
A B
0 a 1
0 a 2
1 b 3
1 b 4
1 b 5
I don't know how good this approach is but it works when you have a list of items.
我不知道这种方法有多好，但是当您有项目列表时它会起作用。
回答by jezrael
Faster solution with chain.from_iterableand numpy.repeat:
使用chain.from_iterable和更快的解决方案numpy.repeat：
from itertools import chain
import numpy as np
import pandas as pd
df = pd.DataFrame({'A':['a','b'],
'B':[[['A1', 'A2']],[['A1', 'A2', 'A3']]]})
print (df)
A B
0 a [[A1, A2]]
1 b [[A1, A2, A3]]
df1 = pd.DataFrame({ "A": np.repeat(df.A.values,
[len(x) for x in (chain.from_iterable(df.B))]),
"B": list(chain.from_iterable(chain.from_iterable(df.B)))})
print (df1)
A B
0 a A1
1 a A2
2 b A1
3 b A2
4 b A3
Timings:
时间：
A = np.unique(np.random.randint(0, 1000, 1000))
B = [[list(string.ascii_letters[:random.randint(3, 10)])] for _ in range(len(A))]
df = pd.DataFrame({"A":A, "B":B})
print (df)
A B
0 0 [[a, b, c, d, e, f, g, h]]
1 1 [[a, b, c]]
2 3 [[a, b, c, d, e, f, g, h, i]]
3 5 [[a, b, c, d, e]]
4 6 [[a, b, c, d, e, f, g, h, i]]
5 7 [[a, b, c, d, e, f, g]]
6 8 [[a, b, c, d, e, f]]
7 10 [[a, b, c, d, e, f]]
8 11 [[a, b, c, d, e, f, g]]
9 12 [[a, b, c, d, e, f, g, h, i]]
10 13 [[a, b, c, d, e, f, g, h]]
...
...
In [67]: %timeit pd.DataFrame({ "A": np.repeat(df.A.values, [len(x) for x in (chain.from_iterable(df.B))]),"B": list(chain.from_iterable(chain.from_iterable(df.B)))})
1000 loops, best of 3: 818 μs per loop
In [68]: %timeit ((df['B'].apply(lambda x: pd.Series(x[0])).stack().reset_index(level=1, drop=True).to_frame('B').join(df[['A']], how='left')))
10 loops, best of 3: 103 ms per loop
回答by Howardyan
I can't find a elegant way to handle this, but the following codes can work...
我找不到一种优雅的方法来处理这个问题，但以下代码可以工作......
import pandas as pd
import numpy as np
df = pd.DataFrame([{"a":1,"b":[[1,2]]},{"a":4, "b":[[3,4,5]]}])
z = []
for k,row in df.iterrows():
for j in list(np.array(row.b).flat):
z.append({'a':row.a, 'b':j})
result = pd.DataFrame(z)
回答by Adrian P
I think this is the fastest and simplest way:
我认为这是最快和最简单的方法：
df = pd.DataFrame({'A':['a','b'],
'B':[[['A1', 'A2']],[['A1', 'A2', 'A3']]]})
df.set_index('A')['B'].apply(lambda x: pd.Series(x[0]))
回答by oli5679
Here's another option
这是另一种选择
unpacked = (pd.melt(df.B.apply(pd.Series).reset_index(),id_vars='index')
.merge(df, left_on = 'index', right_index = True))
unpacked = (unpacked.loc[unpacked.value.notnull(),:]
.drop(columns=['index','variable','B'])
.rename(columns={'value':'B'})
- Apply pd.series to column B --> splits each list entry to a different row
- Melt this, so that each entry is a separate row (preserving index)
- Merge this back on original dataframe
- Tidy up - drop unnecessary columns and rename the values column
- 将 pd.series 应用于 B 列 --> 将每个列表条目拆分为不同的行
- 融化这个，这样每个条目都是一个单独的行（保留索引）
- 将其合并回原始数据帧
- 整理 - 删除不必要的列并重命名值列
