python - 用 NumPy 向量化 groupby

Question

Pandas 有一个广泛使用的 groupby根据相应的映射拆分 DataFrame 的工具，您可以从中对每个子组应用计算并重新组合结果。

这可以在没有原生 Python for 循环的情况下在 NumPy 中灵活地完成吗？使用 Python 循环，这看起来像:

>>> import numpy as np

>>> X = np.arange(10).reshape(5, 2)
>>> groups = np.array([0, 0, 0, 1, 1])

# Split up elements (rows) of `X` based on their element wise group
>>> np.array([X[groups==i].sum() for i in np.unique(groups)])
array([15, 30])

上面15是X前三行的和，30是剩下两行的和。

我所说的“灵活”只是指我们不专注于某个特定的计算，例如求和、计数、最大值等，而是将任何计算传递给分组数组。

如果没有，是否有比上述方法更快的方法？

Answer 1

如何使用 scipy 稀疏矩阵

import numpy as np
from scipy import sparse
import time

x_len = 500000
g_len = 100

X = np.arange(x_len * 2).reshape(x_len, 2)
groups = np.random.randint(0, g_len, x_len)

# original
s = time.time()

a = np.array([X[groups==i].sum() for i in np.unique(groups)])

print(time.time() - s)

# using scipy sparse matrix
s = time.time()

x_sum = X.sum(axis=1)
b = np.array(sparse.coo_matrix(
    (
        x_sum,
        (groups, np.arange(len(x_sum)))
    ),
    shape=(g_len, x_len)
).sum(axis=1)).ravel()

print(time.time() - s)

#compare
print(np.abs((a-b)).sum())

我电脑上的结果

0.15915322303771973
0.012875080108642578
0

快 10 倍以上。

---

更新!

让我们对@Paul Panzer 和@Daniel F 的答案进行基准测试。它只是求和基准。

import numpy as np
from scipy import sparse
import time

# by @Daniel F
def groupby_np(X, groups, axis = 0, uf = np.add, out = None, minlength = 0, identity = None):
    if minlength < groups.max() + 1:
        minlength = groups.max() + 1
    if identity is None:
        identity = uf.identity
    i = list(range(X.ndim))
    del i[axis]
    i = tuple(i)
    n = out is None
    if n:
        if identity is None:  # fallback to loops over 0-index for identity
            assert np.all(np.in1d(np.arange(minlength), groups)), "No valid identity for unassinged groups"
            s = [slice(None)] * X.ndim
            for i_ in i:
                s[i_] = 0
            out = np.array([uf.reduce(X[tuple(s)][groups == i]) for i in range(minlength)])
        else:
            out = np.full((minlength,), identity, dtype = X.dtype)
    uf.at(out, groups, uf.reduce(X, i))
    if n:
        return out

x_len = 500000
g_len = 200

X = np.arange(x_len * 2).reshape(x_len, 2)
groups = np.random.randint(0, g_len, x_len)

print("original")
s = time.time()

a = np.array([X[groups==i].sum() for i in np.unique(groups)])

print(time.time() - s)

print("use scipy coo matrix")
s = time.time()

x_sum = X.sum(axis=1)
b = np.array(sparse.coo_matrix(
    (
        x_sum,
        (groups, np.arange(len(x_sum)))
    ),
    shape=(g_len, x_len)
).sum(axis=1)).ravel()

print(time.time() - s)

#compare
print(np.abs((a-b)).sum())


print("use scipy csr matrix @Daniel F")
s = time.time()
x_sum = X.sum(axis=1)
c = np.array(sparse.csr_matrix(
    (
        x_sum,
        groups,
        np.arange(len(groups)+1)
    ),
    shape=(len(groups), g_len)
).sum(axis=0)).ravel()

print(time.time() - s)

#compare
print(np.abs((a-c)).sum())


print("use bincount @Paul Panzer @Daniel F")
s = time.time()
d = np.bincount(groups, X.sum(axis=1), g_len)
print(time.time() - s)

#compare
print(np.abs((a-d)).sum())

print("use ufunc @Daniel F")
s = time.time()
e = groupby_np(X, groups)
print(time.time() - s)

#compare
print(np.abs((a-e)).sum())

标准输出

original
0.2882847785949707
use scipy coo matrix
0.012301445007324219
0
use scipy csr matrix @Daniel F
0.01046299934387207
0
use bincount @Paul Panzer @Daniel F
0.007468223571777344
0.0
use ufunc @Daniel F
0.04431319236755371
0

获胜者是 bincount 解决方案。但是csr矩阵的解法也很有意思。