[英]Prepend a level to a pandas MultiIndex
我有一个带有 MultiIndex 的 DataFrame 在一些分组后创建:
import numpy as np
import pandas as pd
from numpy.random import randn
df = pd.DataFrame({'A' : ['a1', 'a1', 'a2', 'a3'],
'B' : ['b1', 'b2', 'b3', 'b4'],
'Vals' : randn(4)}
).groupby(['A', 'B']).sum()
# Vals
# A B
# a1 b1 -1.632460
# b2 0.596027
# a2 b3 -0.619130
# a3 b4 -0.002009
如何在 MultiIndex 前面添加一个级别,以便将其转换为以下内容:
# Vals
# FirstLevel A B
# Foo a1 b1 -1.632460
# b2 0.596027
# a2 b3 -0.619130
# a3 b4 -0.002009
使用pandas.concat()
在一行中完成此操作的好方法:
import pandas as pd
pd.concat([df], keys=['Foo'], names=['Firstlevel'])
更短的方法:
pd.concat({'Foo': df}, names=['Firstlevel'])
这可以推广到许多数据框,请参阅文档。
您可以先将其添加为普通列,然后将其附加到当前索引,因此:
df['Firstlevel'] = 'Foo'
df.set_index('Firstlevel', append=True, inplace=True)
并根据需要更改顺序:
df.reorder_levels(['Firstlevel', 'A', 'B'])
结果是:
Vals
Firstlevel A B
Foo a1 b1 0.871563
b2 0.494001
a2 b3 -0.167811
a3 b4 -1.353409
我认为这是一个更通用的解决方案:
# Convert index to dataframe
old_idx = df.index.to_frame()
# Insert new level at specified location
old_idx.insert(0, 'new_level_name', new_level_values)
# Convert back to MultiIndex
df.index = pandas.MultiIndex.from_frame(old_idx)
与其他答案相比的一些优势:
我从cxrodgers answer中创建了一个小函数,恕我直言,这是最好的解决方案,因为它完全适用于索引,独立于任何数据框或系列。
我添加了一个修复: to_frame()
方法将为没有的索引级别发明新名称。 因此,新索引将具有旧索引中不存在的名称。 我添加了一些代码来恢复这个名称更改。
下面是代码,我自己使用了一段时间,它似乎工作正常。 如果您发现任何问题或边缘情况,我将不得不调整我的答案。
import pandas as pd
def _handle_insert_loc(loc: int, n: int) -> int:
"""
Computes the insert index from the right if loc is negative for a given size of n.
"""
return n + loc + 1 if loc < 0 else loc
def add_index_level(old_index: pd.Index, value: Any, name: str = None, loc: int = 0) -> pd.MultiIndex:
"""
Expand a (multi)index by adding a level to it.
:param old_index: The index to expand
:param name: The name of the new index level
:param value: Scalar or list-like, the values of the new index level
:param loc: Where to insert the level in the index, 0 is at the front, negative values count back from the rear end
:return: A new multi-index with the new level added
"""
loc = _handle_insert_loc(loc, len(old_index.names))
old_index_df = old_index.to_frame()
old_index_df.insert(loc, name, value)
new_index_names = list(old_index.names) # sometimes new index level names are invented when converting to a df,
new_index_names.insert(loc, name) # here the original names are reconstructed
new_index = pd.MultiIndex.from_frame(old_index_df, names=new_index_names)
return new_index
它通过了以下单元测试代码:
import unittest
import numpy as np
import pandas as pd
class TestPandaStuff(unittest.TestCase):
def test_add_index_level(self):
df = pd.DataFrame(data=np.random.normal(size=(6, 3)))
i1 = add_index_level(df.index, "foo")
# it does not invent new index names where there are missing
self.assertEqual([None, None], i1.names)
# the new level values are added
self.assertTrue(np.all(i1.get_level_values(0) == "foo"))
self.assertTrue(np.all(i1.get_level_values(1) == df.index))
# it does not invent new index names where there are missing
i2 = add_index_level(i1, ["x", "y"]*3, name="xy", loc=2)
i3 = add_index_level(i2, ["a", "b", "c"]*2, name="abc", loc=-1)
self.assertEqual([None, None, "xy", "abc"], i3.names)
# the new level values are added
self.assertTrue(np.all(i3.get_level_values(0) == "foo"))
self.assertTrue(np.all(i3.get_level_values(1) == df.index))
self.assertTrue(np.all(i3.get_level_values(2) == ["x", "y"]*3))
self.assertTrue(np.all(i3.get_level_values(3) == ["a", "b", "c"]*2))
# df.index = i3
# print()
# print(df)
用pandas.MultiIndex.from_tuples从头开始构建它怎么样?
df.index = p.MultiIndex.from_tuples(
[(nl, A, B) for nl, (A, B) in
zip(['Foo'] * len(df), df.index)],
names=['FirstLevel', 'A', 'B'])
与cxrodger 的解决方案类似,这是一种灵活的方法,可避免修改数据帧的底层数组。
使用from_tuples()
另一个答案。 这概括了之前的答案(请在那里投票)。
key = "Foo"
name = "First"
# If df.index.nlevels > 1:
df.index = pd.MultiIndex.from_tuples(((key, *item) for item in df.index),
names=[name]+df.index.names)
# If df.index.nlevels == 1:
# df.index = pd.MultiIndex.from_tuples(((key, item) for item in df.index),
# names=[name]+df.index.names)
我喜欢这种方法,因为
将上述内容包装在一个函数中可以更轻松地在行和列索引之间以及单级和多级索引之间切换:
def prepend_index_level(index, key, name=None):
names = index.names
if index.nlevels==1:
# Sequence of tuples
index = ((item,) for item in index)
tuples_gen = ((key,)+item for item in index)
return pd.MultiIndex.from_tuples(tuples_gen, names=[name]+names)
df.index = prepend_index_level(df.index, key="Foo", name="First")
df.columns = prepend_index_level(df.columns, key="Bar", name="Top")
# Top Bar
# Vals
# First A B
# Foo a1 b1 -0.446066
# b2 -0.248027
# a2 b3 0.522357
# a3 b4 0.404048
最后,可以通过在任何索引级别插入键来进一步概括上述内容:
def insert_index_level(index, key, name=None, level=0):
def insert_(pos, seq, value):
seq = list(seq)
seq.insert(pos, value)
return tuple(seq)
names = insert_(level, index.names, name)
if index.nlevels==1:
# Sequence of tuples.
index = ((item,) for item in index)
tuples_gen = (insert_(level, item, key) for item in index)
return pd.MultiIndex.from_tuples(tuples_gen, names=names)
df.index = insert_index_level(df.index, key="Foo", name="Last", level=2)
df.columns = insert_index_level(df.columns, key="Bar", name="Top", level=0)
# Top Bar
# Vals
# A B Last
# a1 b1 Foo -0.595949
# b2 Foo -1.621233
# a2 b3 Foo -0.748917
# a3 b4 Foo 2.147814
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