您好我有一个具有2级多索引和一列的DataFrame / Series。我想采用二级索引并将其用作列。例如(代码取自multi-index docs):
import pandas as pd
import numpy as np
arrays = [['bar', 'bar', 'baz', 'baz', 'foo', 'foo', 'qux', 'qux'],
['one', 'two', 'one', 'two', 'one', 'two', 'one', 'two']]
tuples = list(zip(*arrays))
index = pd.MultiIndex.from_tuples(tuples, names=['first', 'second'])
s = pd.DataFrame(np.random.randn(8), index=index, columns=["col"])
看起来像:
first second
bar one -0.982656
two -0.078237
baz one -0.345640
two -0.160661
foo one -0.605568
two -0.140384
qux one 1.434702
two -1.065408
dtype: float64
我希望拥有一个索引为[bar, baz, foo, qux]
和列[one, two]
的数据框。
答案 0 :(得分:16)
你只需要unstack
你的系列剧:
>>> s.unstack(level=1)
second one two
first
bar -0.713374 0.556993
baz 0.523611 0.328348
foo 0.338351 -0.571854
qux 0.036694 -0.161852
答案 1 :(得分:2)
这是使用数组重塑的解决方案 -
>>> idx = s.index.levels
>>> c = len(idx[1])
>>> pd.DataFrame(s.values.reshape(-1,c),index=idx[0].values, columns=idx[1].values)
one two
bar 2.225401 1.624866
baz 1.067359 0.349440
foo -0.468149 -0.352303
qux 1.215427 0.429146
如果您不关心索引顶部显示的名称 -
>>> pd.DataFrame(s.values.reshape(-1,c), index=idx[0], columns=idx[1])
second one two
first
bar 2.225401 1.624866
baz 1.067359 0.349440
foo -0.468149 -0.352303
qux 1.215427 0.429146
给定数据集大小的计时 -
# @AChampion's solution
In [201]: %timeit s.unstack(level=1)
1000 loops, best of 3: 444 µs per loop
# Using array reshaping step-1
In [199]: %timeit s.index.levels
1000000 loops, best of 3: 214 ns per loop
# Using array reshaping step-2
In [202]: %timeit pd.DataFrame(s.values.reshape(-1,2), index=idx[0], columns=idx[1])
10000 loops, best of 3: 47.3 µs per loop
答案 2 :(得分:1)
另一个强大的解决方案是使用.reset_index
和.pivot
:
levels= [['bar', 'baz'], ['one', 'two', 'three']]
index = pd.MultiIndex.from_product(levels, names=['first', 'second'])
series = pd.Series(np.random.randn(6), index)
df = series.reset_index()
# Shorthand notation instead of explicitly naming index, columns and values
df = df.pivot(*df.columns)
结果:
second one three two
first
bar 1.047692 1.209063 0.891820
baz 0.083602 -0.303528 -1.385458