将numpy.digitize扩展为多维数据

时间:2014-07-08 23:40:34

标签: python numpy

我有一组大型数组(每个大约600万个元素),我想基本上执行np.digitize但是多个轴。我正在寻找有关如何有效地执行此操作以及如何存储结果的一些建议。

我需要数组A的所有索引(或所有值或掩码),其中数组B的值在一个范围内,而数组C的值在另一个范围内,D在另一个范围内。我想要值,索引或掩码,以便我可以做一些关于每个bin中A数组值的尚未确定的统计信息。我还需要每个bin中的元素数量,但len()可以做到这一点。

这是我编写的一个看似合理的例子:

import itertools
import numpy as np

A = np.random.random_sample(1e4)
B = (np.random.random_sample(1e4) + 10)*20
C = (np.random.random_sample(1e4) + 20)*40
D = (np.random.random_sample(1e4) + 80)*80

# make the edges of the bins
Bbins = np.linspace(B.min(), B.max(), 10)
Cbins = np.linspace(C.min(), C.max(), 12) # note different number
Dbins = np.linspace(D.min(), D.max(), 24) # note different number

B_Bidx = np.digitize(B, Bbins)
C_Cidx = np.digitize(C, Cbins)
D_Didx = np.digitize(D, Dbins)

a_bins = []
for bb, cc, dd in itertools.product(np.unique(B_Bidx), 
                                    np.unique(C_Cidx), 
                                    np.unique(D_Didx)):
    a_bins.append([(bb, cc, dd), [A[np.bitwise_and((B_Bidx==bb),
                                                   (C_Cidx==cc),
                                                   (D_Didx==dd))]]])

然而,这使我感到紧张,我将在大型阵列上耗尽内存。

我也可以这样做:

b_inds = np.empty((len(A), 10), dtype=np.bool)
c_inds = np.empty((len(A), 12), dtype=np.bool)
d_inds = np.empty((len(A), 24), dtype=np.bool)
for i in range(10):
    b_inds[:,i] = B_Bidx = i     
for i in range(12):
    c_inds[:,i] = C_Cidx = i     
for i in range(24):
    d_inds[:,i] = D_Didx = i     
# get the A data for the 1,2,3 B,C,D bin
print A[b_inds[:,1] & c_inds[:,2] & d_inds[:,3]]

至少在这里输出是已知且恒定的。

有没有人对如何更聪明地做出更好的想法?还是需要澄清?


根据HYRY的答案,这是我决定采取的路径。

import numpy as np
import pandas as pd

np.random.seed(42)
A =  np.random.random_sample(1e7)
B = (np.random.random_sample(1e7) + 10)*20
C = (np.random.random_sample(1e7) + 20)*40
D = (np.random.random_sample(1e7) + 80)*80
# make the edges of the bins we want
Bbins = np.linspace(B.min(), B.max(), 9)
Cbins = np.linspace(C.min(), C.max(), 10) # note different number
Dbins = np.linspace(D.min(), D.max(), 11) # note different number
sA = pd.Series(A)
cB = pd.cut(B, Bbins, include_lowest=True)
cC = pd.cut(C, Cbins, include_lowest=True)
cD = pd.cut(D, Dbins, include_lowest=True)

dat = pd.DataFrame({'A':A, 'cB':cB.labels, 'cC':cC.labels, 'cD':cD.labels})
g = sA.groupby([cB.labels, cC.labels, cD.labels]).indices
# this then gives all the indices that match the group 
print g[0,1,2]
# this is all the array A data for that B,C,D bin
print sA[g[0,1,2]]

即使对于大型阵列,这种方法看起来也很快。

1 个答案:

答案 0 :(得分:5)

如何在Pandas中使用groupby。首先修复代码中的一些问题:

import itertools
import numpy as np

np.random.seed(42)

A = np.random.random_sample(1e4)
B = (np.random.random_sample(1e4) + 10)*20
C = (np.random.random_sample(1e4) + 20)*40
D = (np.random.random_sample(1e4) + 80)*80

# make the edges of the bins
Bbins = np.linspace(B.min(), B.max(), 10)
Cbins = np.linspace(C.min(), C.max(), 12) # note different number
Dbins = np.linspace(D.min(), D.max(), 24) # note different number

B_Bidx = np.digitize(B, Bbins)
C_Cidx = np.digitize(C, Cbins)
D_Didx = np.digitize(D, Dbins)

a_bins = []
for bb, cc, dd in itertools.product(np.unique(B_Bidx), 
                                    np.unique(C_Cidx), 
                                    np.unique(D_Didx)):
    a_bins.append([(bb, cc, dd), A[(B_Bidx==bb) & (C_Cidx==cc) & (D_Didx==dd)]])

a_bins[1000]

输出:

[(4, 6, 17), array([ 0.70723863,  0.907611  ,  0.46214047])]

以下是Pandas返回相同结果的代码:

import pandas as pd

cB = pd.cut(B, 9)
cC = pd.cut(C, 11)
cD = pd.cut(D, 23)

sA = pd.Series(A)
g = sA.groupby([cB.labels, cC.labels, cD.labels])
g.get_group((3, 5, 16))

输出:

800     0.707239
2320    0.907611
9388    0.462140
dtype: float64

如果要计算每个组的某些统计信息,可以调用g的方法,例如:

g.mean()

返回:

0  0  0     0.343566
      1     0.410979
      2     0.700007
      3     0.189936
      4     0.452566
      5     0.565330
      6     0.539565
      7     0.530867
      8     0.568120
      9     0.587762
      11    0.352453
      12    0.484903
      13    0.477969
      14    0.484328
      15    0.467357
...
8  10  8     0.559859
       9     0.570652
       10    0.656718
       11    0.353938
       12    0.628980
       13    0.372350
       14    0.404543
       15    0.387920
       16    0.742292
       17    0.530866
       18    0.389236
       19    0.628461
       20    0.387384
       21    0.541831
       22    0.573023
Length: 2250, dtype: float64