创建我的数据框:
from pandas import *
arrays = [['bar', 'bar', 'baz', 'baz', 'foo', 'foo', 'qux', 'qux'],
['one', 'two', 'one', 'two', 'one', 'two', 'one', 'two']]
tuples = zip(*arrays)
index = MultiIndex.from_tuples(tuples, names=['first','second'])
data = DataFrame(randn(8,2),index=index,columns=['c1','c2'])
data
Out[68]:
c1 c2
first second
bar one 0.833816 -1.529639
two 0.340150 -1.818052
baz one -1.605051 -0.917619
two -0.021386 -0.222951
foo one 0.143949 -0.406376
two 1.208358 -2.469746
qux one -0.345265 -0.505282
two 0.158928 1.088826
我想重命名“第一个”索引值,例如“bar” - >“cat”,“baz” - >“dog”等。但是,我读过的每个例子都在单级索引和/或循环遍历整个索引,从头开始有效地重新创建它。我想的是:
data = data.reindex(index={'bar':'cat','baz':'dog'})
但是这不起作用,我也不希望它在多个索引上工作。我可以在不循环整个数据框索引的情况下进行这样的替换吗?
开始修改
我对在发布之前更新到0.13犹豫不决,所以我使用了以下解决方法:
index = data.index.tolist()
for r in xrange( len(index) ):
index[r] = (codes[index[r][0]],index[r][1])
index = pd.MultiIndex.from_tuples(index,names=data.index.names)
data.index = index
以前定义的代码字典在哪里:字符串对。这实际上并没有像我期望的那样大的性能(需要几秒钟来操作~110万行)。它不像单线,但确实有效。
结束修改
答案 0 :(得分:16)
使用set_levels
方法(new in version 0.13.0):
data.index.set_levels([[u'cat', u'dog', u'foo', u'qux'],
[u'one', u'two']], inplace=True)
产量
c1 c2
first second
cat one -0.289649 -0.870716
two -0.062014 -0.410274
dog one 0.030171 -1.091150
two 0.505408 1.531108
foo one 1.375653 -1.377876
two -1.478615 1.351428
qux one 1.075802 0.532416
two 0.865931 -0.765292
要根据dict重新映射级别,可以使用如下函数:
def map_level(df, dct, level=0):
index = df.index
index.set_levels([[dct.get(item, item) for item in names] if i==level else names
for i, names in enumerate(index.levels)], inplace=True)
dct = {'bar':'cat', 'baz':'dog'}
map_level(data, dct, level=0)
这是一个可运行的例子:
import numpy as np
import pandas as pd
arrays = [['bar', 'bar', 'baz', 'baz', 'foo', 'foo', 'qux', 'qux'],
['one', 'two', 'one', 'two', 'one', 'two', 'one', 'two']]
tuples = zip(*arrays)
index = pd.MultiIndex.from_tuples(tuples, names=['first','second'])
data = pd.DataFrame(np.random.randn(8,2),index=index,columns=['c1','c2'])
data2 = data.copy()
data.index.set_levels([[u'cat', u'dog', u'foo', u'qux'],
[u'one', u'two']], inplace=True)
print(data)
# c1 c2
# first second
# cat one 0.939040 -0.748100
# two -0.497006 -1.185966
# dog one -0.368161 0.050339
# two -2.356879 -0.291206
# foo one -0.556261 0.474297
# two 0.647973 0.755983
# qux one -0.017722 1.364244
# two 1.007303 0.004337
def map_level(df, dct, level=0):
index = df.index
index.set_levels([[dct.get(item, item) for item in names] if i==level else names
for i, names in enumerate(index.levels)], inplace=True)
dct = {'bar':'wolf', 'baz':'rabbit'}
map_level(data2, dct, level=0)
print(data2)
# c1 c2
# first second
# wolf one 0.939040 -0.748100
# two -0.497006 -1.185966
# rabbit one -0.368161 0.050339
# two -2.356879 -0.291206
# foo one -0.556261 0.474297
# two 0.647973 0.755983
# qux one -0.017722 1.364244
# two 1.007303 0.004337
答案 1 :(得分:1)
set_levels
方法导致我的新列名无序。所以我找到了一个不太干净的不同解决方案,但效果很好。方法是print df.index
(或等效df.columns
),然后复制并粘贴输出,并更改所需的值。例如:
print data.index
MultiIndex(等级= [[' bar',' baz',' foo',' qux'],[&#39 ;一个','两个']],标签= [[0,0,1,1,2,2,3,3],[0,1,0,1,0,1] ,0,1]], names = [' first',' second'])
data.index = MultiIndex(levels=[['new_bar', 'new_baz', 'new_foo', 'new_qux'],
['new_one', 'new_two']],
labels=[[0, 0, 1, 1, 2, 2, 3, 3], [0, 1, 0, 1, 0, 1, 0, 1]],
names=['first', 'second'])
我们也可以通过编辑标签来完全控制名称。例如:
data.index = MultiIndex(levels=[['bar', 'baz', 'foo', 'qux'],
['one', 'twooo', 'three', 'four',
'five', 'siz', 'seven', 'eit']],
labels=[[0, 0, 1, 1, 2, 2, 3, 3], [0, 1, 2, 3, 4, 5, 6, 7]],
names=['first', 'second'])
请注意,在此示例中,我们已经完成了from pandas import MultiIndex
或from pandas import *
之类的操作。