Nic*_*mer 5 python arrays numpy
我有一个(大)整数数组,如
materials = [0, 0, 47, 0, 2, 2, 47] # ...
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几乎没有独特的条目,我想将其转换为索引字典,即,
d = {
0: [0, 1, 3],
2: [4, 5],
47: [2, 6],
}
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这样做的最有效方法是什么?(欢迎使用 NumPy。)
另一句俏皮话,这次是numpy.where:
out = {v: np.where(v == a)[0] for v in numpy.unique(a)}
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(对于某些应用程序,布尔数组可能就足够了:
out = {v: v == a for v in numpy.unique(a)}
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)
请注意,这numpy.unique比大型数组要快set(),并且如果只有几个唯一条目,则速度要快很多。
无论如何,对于大多数数组大小,上面的方法是迄今为止最快的方法:
代码:
import numpy as np
from collections import defaultdict
import perfplot
def pp(a):
index = np.argsort(a, kind='mergesort')
as_ = a[index]
jumps = np.r_[0, 1 + np.where(np.diff(as_) != 0)[0]]
pp_out = {k: v for k, v in zip(as_[jumps], np.split(index, jumps[1:]))}
return pp_out
def pp2(a):
index = np.argsort(a)
as_ = a[index]
jumps = np.r_[0, 1 + np.where(np.diff(as_) != 0)[0]]
pp_out = {k: np.sort(v)
for k, v in zip(as_[jumps], np.split(index, jumps[1:]))}
return pp_out
def Denziloe_JFFabre(a):
df_out = {v: [i for i, x in enumerate(a) if x == v] for v in np.unique(a)}
return df_out
def FCouzo(a):
fc_out = defaultdict(list)
for i, elem in enumerate(a):
fc_out[elem].append(i)
return fc_out
def KKSingh(a):
kks_out = defaultdict(list)
list(map(lambda x: kks_out[x[0]].append(x[1]), zip(a, range(len(a)))))
return kks_out
def TMcDonaldJensen(a):
mdj_out = defaultdict(list)
for i, elem in enumerate(a):
mdj_out[elem].append(i)
return mdj_out
def RomanPerekhrest(a):
rp_out = {}
for k, m in enumerate(a):
rp_out.setdefault(m, []).append(k)
return rp_out
def SchloemerHist(a):
np.histogram(a, bins=np.arange(min(a), max(a)+2))
return
def SchloemerWhere(a):
out = {v: np.where(v == a)[0] for v in np.unique(a)}
return out
def SchloemerBooleanOnly(a):
out = {v: v == a for v in np.unique(a)}
return out
perfplot.show(
setup=lambda n: np.random.randint(0, 100, n),
kernels=[
pp, pp2, Denziloe_JFFabre, FCouzo, KKSingh,
TMcDonaldJensen, RomanPerekhrest, SchloemerHist, SchloemerWhere,
SchloemerBooleanOnly
],
n_range=[2**k for k in range(17)],
xlabel='len(a)',
logx=True,
logy=True,
)
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