我不會(huì)說(shuō)這是重復(fù)的，但您提到的相關(guān)問(wèn)題是一個(gè)很好的起點(diǎn)。您鏈接的大多數(shù)答案都需要對(duì)數(shù)組進(jìn)行排序，提取組開始的索引，然后調(diào)用np.split它。這里不是這種情況，因?yàn)樗鼤?huì)返回一個(gè)大小不平衡的組列表。
相反，您可以使用np.bincount方法。它計(jì)算每個(gè)加權(quán)值出現(xiàn)的次數(shù)，這實(shí)際上與 groupby sum 相同，只是輸出中缺少組鍵。
def group_by_sum(x):
u, idx = np.unique(x[:,0], return_inverse=True)
s = np.bincount(idx, weights = x[:,1])
return np.c_[u, s]
獎(jiǎng)金。它實(shí)際上是numpy_indexed包裝中的一個(gè)單行：
np.transpose(npi.group_by(x[:, 0]).sum(x[:, 1]))
標(biāo)桿管理
import numpy as np
import perfplot
import matplotlib.pyplot as plt
def bincount(x):
u, idx = np.unique(x[:,0], return_inverse=True)
s = np.bincount(idx, weights = x[:,1])
return np.c_[u, s]
def reduceat(x):
x = x[np.argsort(x[:, 0])]
i = np.flatnonzero(np.diff(x[:, 0]))
i = np.r_[0, i + 1]
s = np.add.reduceat(x[:, 1], i)
return np.stack((x[i, 0], s), axis=-1)
def setup(N, s):
x = np.linspace(0,1,N+1)[np.random.randint(N, size = s)]
return np.c_[x, (x**2)%1]
def build_args(k):
return {'setup': lambda x: setup(k, x),
'kernels': [bincount, reduceat],
'n_range': [2**k for k in range(1, 20)],
'title': f'Testing for x samples in [0, 1] with no more than {k} groups',
'show_progress': True,
'equality_check': False}
outs = [perfplot.bench(**build_args(n)) for n in (10, 100, 1000, 10000)]
fig = plt.figure(figsize=(20, 20))
for i in range(len(outs)):
ax = fig.add_subplot(2, 2, i + 1)
ax.grid(True, which="both")
outs[i].plot()
plt.show()
