如何恢复扁平Numpy数组的原始索引？

Question

I've got a multidimensional numpy array that I'm trying to stick into a pandas data frame. I'd like to flatten the array, and create a pandas index that reflects the pre-flattened array indices.

Note I'm using 3D to keep the example small, but I'd like to generalize to at least 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())

I'm trying to generate x/y/z columns like this:

           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

I've tried setting this up using np.meshgrid but I'm going wrong somewhere:

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()

This result looks somewhat sensible, but the indices are wrong:

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

How can I create the index columns to reflect the shape of A ?

Answer 1

You could use 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'))

yields

    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

Or if performance is a top priority, consider using senderle's cartesian_product . (See the code, below.)

---

Here is a benchmark for A with shape (100, 100, 100):

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

---

This is the setup I used for the benchmark:

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

def indices_merged_arr_generic(arr, columns):
    n = arr.ndim
    grid = np.ogrid[tuple(map(slice, arr.shape))]
    out = np.empty(arr.shape + (n+1,), dtype=arr.dtype)
    for i in range(n):
        out[...,i] = grid[i]
    out[...,-1] = arr
    out.shape = (-1,n+1)
    df = pd.DataFrame(out, columns=['A']+columns)
    return df

def using_product(A):
    x, y, z = A.shape
    x_, y_, z_ = zip(*product(range(x), range(y), range(z)))
    df = pd.DataFrame(A.flatten()).assign(x=x_, y=y_, z=z_)
    return df

A = np.random.random((100,100,100))
shape = A.shape
columns = list(string.ascii_uppercase[-len(shape):][::-1])

Answer 2

from itertools import product

np.random.seed(0)
A = np.random.rand(2, 3, 4)
x, y, z = A.shape
x_, y_, z_ = zip(*product(range(x), range(y), range(z)))
df = pd.DataFrame(A.flatten()).assign(x=x_, y=y_, z=z_)
>>> df

           0  x  y  z
0   0.548814  0  0  0
1   0.715189  0  0  1
2   0.602763  0  0  2
3   0.544883  0  0  3
4   0.423655  0  1  0
5   0.645894  0  1  1
6   0.437587  0  1  2
7   0.891773  0  1  3
8   0.963663  0  2  0
9   0.383442  0  2  1
10  0.791725  0  2  2
11  0.528895  0  2  3
12  0.568045  1  0  0
13  0.925597  1  0  1
14  0.071036  1  0  2
15  0.087129  1  0  3
16  0.020218  1  1  0
17  0.832620  1  1  1
18  0.778157  1  1  2
19  0.870012  1  1  3
20  0.978618  1  2  0
21  0.799159  1  2  1
22  0.461479  1  2  2
23  0.780529  1  2  3

Answer 3

My solution is based on this this answer by Divakar involving np.ogrid . This function should work for any array of any dimension.

def indices_merged_arr(arr):
    n = arr.ndim
    grid = np.ogrid[tuple(map(slice, arr.shape))]
    out = np.empty(arr.shape + (n+1,), dtype=arr.dtype)
    for i in range(n):
        out[...,i+1] = grid[i]
    out[...,0] = arr
    out.shape = (-1,n+1)
    return out

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 = pd.DataFrame(indices_merged_arr(A), columns=list('Axyz'))
df

           A    x    y    z
0   0.437939  0.0  0.0  0.0
1   0.400781  0.0  0.0  1.0
2   0.480787  0.0  0.0  2.0
3   0.053342  0.0  0.0  3.0
4   0.763315  0.0  1.0  0.0
5   0.825144  0.0  1.0  1.0
6   0.861691  0.0  1.0  2.0
7   0.864961  0.0  1.0  3.0
8   0.755727  0.0  2.0  0.0
9   0.808609  0.0  2.0  1.0
10  0.799953  0.0  2.0  2.0
11  0.631237  0.0  2.0  3.0
12  0.206489  1.0  0.0  0.0
13  0.570423  1.0  0.0  1.0
14  0.717773  1.0  0.0  2.0
15  0.341550  1.0  0.0  3.0
16  0.308437  1.0  1.0  0.0
17  0.393814  1.0  1.0  1.0
18  0.126235  1.0  1.0  2.0
19  0.934816  1.0  1.0  3.0
20  0.326777  1.0  2.0  0.0
21  0.640970  1.0  2.0  1.0
22  0.304052  1.0  2.0  2.0
23  0.577266  1.0  2.0  3.0

Answer 4

As hpaulj pointed out in a comment, I could add indexing=='ij' to the meshgrid call:

A = np.random.rand(2,3,4)
dimnames = ['z', 'y', 'x']
ranges   = [ np.arange(x) for x in A.shape ]
ix       = [ x.flatten()  for x in np.meshgrid(*ranges, indexing='ij') ]
for name, col in zip(dimnames, ix):
    df[name] = col
df = df.set_index(dimnames).squeeze()

# Compare the results
for ix, val in df.iteritems():
    print ix, val == A[ix]
(0, 0, 0) True
(0, 0, 1) True
(0, 0, 2) True
(0, 0, 3) True
(0, 1, 0) True
(0, 1, 1) True
(0, 1, 2) True
(0, 1, 3) True
(0, 2, 0) True
(0, 2, 1) True
(0, 2, 2) True
(0, 2, 3) True
(1, 0, 0) True
(1, 0, 1) True
(1, 0, 2) True
(1, 0, 3) True
(1, 1, 0) True
(1, 1, 1) True
(1, 1, 2) True
(1, 1, 3) True
(1, 2, 0) True
(1, 2, 1) True
(1, 2, 2) True
(1, 2, 3) True

Answer 5

Another possibility, although others maybe faster...

x,y,z = np.indices(A.shape)

df = pd.DataFrame(np.array([p.flatten() for p in [x,y,z,A]]).T
                  ,columns=['x','y','z',0])

Answer 6

def ndarray_to_indexed_2d(data):
    idx = np.column_stack(np.unravel_index(np.arange(np.product(data.shape[:-1])), data.shape[:-1]))
    data2d = np.hstack((idx, data.reshape(np.product(data.shape[:-1]), data.shape[-1])))
    return data2d