我有一个多维numpy数组，试图将其粘贴到熊猫数据框中。我想展平数组，并创建一个反映预展平数组索引的熊猫索引。
请注意，我使用3D来缩小示例，但我想将其推广到至少4D
A = np.random.rand(2,3,4) array([[[ 0.43793885, 0.40078139, 0.48078691, 0.05334248], [ 0.76331509, 0.82514441, 0.86169078, 0.86496111], [ 0.75572665, 0.80860943, 0.79995337, 0.63123724]], [[ 0.20648946, 0.57042315, 0.71777265, 0.34155005], [ 0.30843717, 0.39381407, 0.12623462, 0.93481552], [ 0.3267771 , 0.64097038, 0.30405215, 0.57726629]]]) df = pd.DataFrame(A.flatten())
我正在尝试生成像这样的x / y / z列：
A z y x 0 0.437939 0 0 0 1 0.400781 0 0 1 2 0.480787 0 0 2 3 0.053342 0 0 3 4 0.763315 0 1 0 5 0.825144 0 1 1 6 0.861691 0 1 2 7 0.864961 0 1 3 ... 21 0.640970 1 2 1 22 0.304052 1 2 2 23 0.577266 1 2 3
我尝试使用进行设置，np.meshgrid但在某处出错：
dimnames = ['z', 'y', 'x'] ranges = [ np.arange(x) for x in A.shape ] ix = [ x.flatten() for x in np.meshgrid(*ranges) ] for name, col in zip(dimnames, ix): df[name] = col df = df.set_index(dimnames).squeeze()
这个结果看起来有些明智，但是索引是错误的：
df z y x 0 0 0 0.437939 1 0.400781 2 0.480787 3 0.053342 1 0 0 0.763315 1 0.825144 2 0.861691 3 0.864961 0 1 0 0.755727 1 0.808609 2 0.799953 3 0.631237 1 1 0 0.206489 1 0.570423 2 0.717773 3 0.341550 0 2 0 0.308437 1 0.393814 2 0.126235 3 0.934816 1 2 0 0.326777 1 0.640970 2 0.304052 3 0.577266 print A[0,1,0] 0.76331508999999997 print print df.loc[0,1,0] 0.75572665000000006
如何创建索引列以反映的形状A？
A
您可以使用pd.MultiIndex.from_product：
import numpy as np import pandas as pd import string def using_multiindex(A, columns): shape = A.shape index = pd.MultiIndex.from_product([range(s)for s in shape], names=columns) df = pd.DataFrame({'A': A.flatten()}, index=index).reset_index() return df A = np.array([[[ 0.43793885, 0.40078139, 0.48078691, 0.05334248], [ 0.76331509, 0.82514441, 0.86169078, 0.86496111], [ 0.75572665, 0.80860943, 0.79995337, 0.63123724]], [[ 0.20648946, 0.57042315, 0.71777265, 0.34155005], [ 0.30843717, 0.39381407, 0.12623462, 0.93481552], [ 0.3267771 , 0.64097038, 0.30405215, 0.57726629]]]) df = using_multiindex(A, list('ZYX'))
产量
Z Y X A 0 0 0 0 0.437939 1 0 0 1 0.400781 2 0 0 2 0.480787 3 0 0 3 0.053342 ... 21 1 2 1 0.640970 22 1 2 2 0.304052 23 1 2 3 0.577266
或者，如果性能是重中之重，请考虑使用senderle的cartesian_product。（请参见下面的代码。）
这是形状为（100，100，100）的A的基准：
In [321]: %timeit using_cartesian_product(A, columns) 100 loops, best of 3: 13.8 ms per loop In [318]: %timeit using_multiindex(A, columns) 10 loops, best of 3: 35.6 ms per loop In [320]: %timeit indices_merged_arr_generic(A, columns) 10 loops, best of 3: 29.1 ms per loop In [319]: %timeit using_product(A) 1 loop, best of 3: 461 ms per loop
这是我用于基准测试的设置：
import numpy as np import pandas as pd import functools import itertools as IT import string product = IT.product def cartesian_product_broadcasted(*arrays): """ http://stackoverflow.com/a/11146645/190597 (senderle) """ broadcastable = np.ix_(*arrays) broadcasted = np.broadcast_arrays(*broadcastable) dtype = np.result_type(*arrays) rows, cols = functools.reduce(np.multiply, broadcasted[0].shape), len(broadcasted) out = np.empty(rows * cols, dtype=dtype) start, end = 0, rows for a in broadcasted: out[start:end] = a.reshape(-1) start, end = end, end + rows return out.reshape(cols, rows).T def using_cartesian_product(A, columns): shape = A.shape coords = cartesian_product_broadcasted(*[np.arange(s, dtype='int') for s in shape]) df = pd.DataFrame(coords, columns=columns) df['A'] = A.flatten() return df def using_multiindex(A, columns): shape = A.shape index = pd.MultiIndex.from_product([range(s)for s in shape], names=columns) df = pd.DataFrame({'A': A.flatten()}, index=index).reset_index() return df
