Question

我有一个多指数df s：

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'])
pd.MultiIndex(levels=[['bar', 'baz', 'foo', 'qux'], ['one', 'two']],
       labels=[[0, 0, 1, 1, 2, 2, 3, 3], [0, 1, 0, 1, 0, 1, 0, 1]],
       names=['first', 'second'])
s = pd.Series(np.random.randn(8), index=index)
s

我想添加一个新的索引列＆＃34;零＆＃34;通过匹配索引列＆＃34;第一个＆＃34;使用x，y，z到s和＆＃34;第二＆＃34;。换句话说，我想重复三次，但是这个额外的索引列有x，y，z。我尝试了reindex（见下文），但为什么它给了我所有的NaN？

mux=pd.MultiIndex.from_product([["x","y","z"], 
                            s.index.get_level_values(0),
                            s.index.get_level_values(1)],
                           names=["zero","first", "second"])
t=s.reindex(mux)
t

我也尝试将匹配级别指定为＆＃34; first＆＃34;和＆＃34;秒＆＃34;，但看起来级别只需要一个整数？

Answer 1

您可以使用reindex，但必须通过MultiIndex创建levels。但它会为现有内容添加新级别，因此如有必要，请添加reorder_levels和sort_index：

np.random.seed(123)
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.Series(np.random.randn(8), index=index)
#print (s)

mux=pd.MultiIndex.from_product([s.index.levels[0],s.index.levels[1], ["x","y","z"]])
t=s.reindex(mux, method='ffill').reorder_levels([2,0,1]).sort_index()
print (t)
x  bar  one   -1.085631
        two    0.997345
   baz  one    0.282978
        two   -1.506295
   foo  one   -0.578600
        two    1.651437
   qux  one   -2.426679
        two   -0.428913
y  bar  one   -1.085631
        two    0.997345
   baz  one    0.282978
        two   -1.506295
   foo  one   -0.578600
        two    1.651437
   qux  one   -2.426679
        two   -0.428913
z  bar  one   -1.085631
        two    0.997345
   baz  one    0.282978
        two   -1.506295
   foo  one   -0.578600
        two    1.651437
   qux  one   -2.426679
        two   -0.428913
dtype: float64

Answer 2

IIUC，您想要pd.concat吗？

s = pd.concat([s] * 3, axis=0, keys=['x', 'y', 'z'])

如果需要，重命名轴：

s = s.rename_axis(['zero', 'first', 'second'])

s 

zero  first  second
x     bar    one       0.510567
             two       0.066620
      baz    one       0.667948
             two      -1.471894
      foo    one       1.881198
             two       0.143628
      qux    one       1.108174
             two      -0.978112
y     bar    one       0.510567
             two       0.066620
      baz    one       0.667948
             two      -1.471894
      foo    one       1.881198
             two       0.143628
      qux    one       1.108174
             two      -0.978112
z     bar    one       0.510567
             two       0.066620
      baz    one       0.667948
             two      -1.471894
      foo    one       1.881198
             two       0.143628
      qux    one       1.108174
             two      -0.978112
dtype: float64

添加索引列并通过匹配部分索引标签重新索引数据帧

2 个答案: