由多个向量组成的Numpy组，获得组索引

Question

I have several numpy arrays; I want to build a groupby method that would have group ids for these arrays. It will then allow me to index these arrays on the group id to perform operations on the groups.

For an example:

import numpy as np
import pandas as pd
a = np.array([1,1,1,2,2,3])
b = np.array([1,2,2,2,3,3])

def group_np(groupcols):
    groupby = np.array([''.join([str(b) for b in bs]) for bs in zip(*[c for c in groupcols])])
    _, groupby = np.unique(groupby, return_invesrse=True)
   return groupby

def group_pd(groupcols):
    df = pd.DataFrame(groupcols[0])
    for i in range(1, len(groupcols)):
        df[i] = groupcols[i]
    for i in range(len(groupcols)):
        df[i] = df[i].fillna(-1)
    return df.groupby(list(range(len(groupcols)))).grouper.group_info[0]

Outputs:

group_np([a,b]) -> [0, 1, 1, 2, 3, 4]
group_pd([a,b]) -> [0, 1, 1, 2, 3, 4]

Is there a more efficient way of implementing it, ideally in pure numpy? The bottleneck currently seems to be building a vector that would have unique values for each group - at the moment I am doing that by concatenating the values for each vector as strings.

I want this to work for any number of input vectors, which can have millions of elements.

Edit: here is another testcase:

a = np.array([1,2,1,1,1,2,3,1])
b = np.array([1,2,2,2,2,3,3,2])

Here, group elements 2,3,4,7 should all be the same.

Edit2: adding some benchmarks.

a = np.random.randint(1, 1000, 30000000)
b = np.random.randint(1, 1000, 30000000)
c = np.random.randint(1, 1000, 30000000)

def group_np2(groupcols):
    _, groupby = np.unique(np.stack(groupcols), return_inverse=True, axis=1)
    return groupby

%timeit group_np2([a,b,c])
# 25.1 s +/- 1.06 s per loop (mean +/- std. dev. of 7 runs, 1 loop each)
%timeit group_pd([a,b,c])
# 21.7 s +/- 646 ms per loop (mean +/- std. dev. of 7 runs, 1 loop each)

Answer 1

After using np.stack on the arrays a and b , if you set the parameter return_inverse to True in np.unique then it is the output you are looking for:

a = np.array([1,2,1,1,1,2,3,1])
b = np.array([1,2,2,2,2,3,3,2])
_, inv = np.unique(np.stack([a,b]), axis=1, return_inverse=True)
print (inv)

array([0, 2, 1, 1, 1, 3, 4, 1], dtype=int64)

and you can replace [a,b] in np.stack by a list of all the vectors.

Edit : a faster solution is use np.unique on the sum of the arrays multiply by the cumulative product ( np.cumprod ) of the max plus 1 of all previous arrays in groupcols . such as:

def group_np_sum(groupcols):
    groupcols_max = np.cumprod([ar.max()+1 for ar in groupcols[:-1]])
    return np.unique( sum([groupcols[0]] +
                          [ ar*m for ar, m in zip(groupcols[1:],groupcols_max)]), 
                      return_inverse=True)[1]

To check:

a = np.array([1,2,1,1,1,2,3,1])
b = np.array([1,2,2,2,2,3,3,2])
print (group_np_sum([a,b]))
array([0, 2, 1, 1, 1, 3, 4, 1], dtype=int64)

Note: the number associated to each group may not be the same (here I changed the first element of a by 3)

a = np.array([3,2,1,1,1,2,3,1])
b = np.array([1,2,2,2,2,3,3,2])
print(group_np2([a,b]))
print (group_np_sum([a,b]))
array([3, 1, 0, 0, 0, 2, 4, 0], dtype=int64)
array([0, 2, 1, 1, 1, 3, 4, 1], dtype=int64)

but groups themselves are the same.

Now to check for timing:

a = np.random.randint(1, 100, 30000)
b = np.random.randint(1, 100, 30000)
c = np.random.randint(1, 100, 30000)
groupcols = [a,b,c]

%timeit group_pd(groupcols)
#13.7 ms ± 1.22 ms per loop (mean ± std. dev. of 7 runs, 100 loops each)

%timeit group_np2(groupcols)
#34.2 ms ± 6.88 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

%timeit group_np_sum(groupcols)
#3.63 ms ± 562 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

Answer 2

The numpy_indexed package (dsiclaimer: I am its authos) covers these type of use cases:

import numpy_indexed as npi
npi.group_by((a, b))

Passing a tuple of index-arrays like this avoids creating a copy; but if you dont mind making the copy you can use stacking as well:

npi.group_by(np.stack(a, b))