关于python：从numpy数组创建值频率字典

Create dictionary of value frequencies from numpy array

我需要从一个循环中创建一个字典，该循环贯穿两列数字的数组。下面是数组的一个子集：

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array([[ 0, 1],
[ 1, 0],
[ 1, 2],
[ 2, 3],
[ 2, 1]])

我想创建一个字典，它将第一列的唯一数字作为键(例如本例中的0,1,2)，将第二列的相应数字作为值。

对于本例，字典如下所示：

1	dict = {0:[1], 1:[0,2], 2:[3,1]}

号

我的数组非常长(370000 x 2)，所以我想通过一个有效的循环来完成这项工作。任何建议都将不胜感激！

您可以使用defaultdict来完成此操作。

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from collections import defaultdict
a = np.array([[ 0, 1],[ 1, 0],[ 1, 2],[ 2, 3], [ 2, 1]])
d = defaultdict(list)
for x,y in a:
d[x].append(y)

相关讨论

纯麻木的解决方案：

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b=a[np.lexsort(a.T[::-1])] # if necessary.
keys,values=b.T
uniq,steps=np.unique(keys,return_index =True)
bins=np.split(values,steps[1:])

号

如果是uniq==range(len(uniq))，那就这样吧：bins[key]会起作用，而且是最快的方法。

其他：

1 2	d=dict(zip(uniq,bins)) #{0: array([1]), 1: array([0, 2]), 2: array([1, 3])}

号

将建立您的字典。

如果第一列是"带重复的范围"

1 2	steps_at = np.searchsorted(a[:,0], np.arange(a[-1,0]+1)) result = {k:v for k,v in zip(a[steps_at,0], np.split(a[:,1], steps_at[1:]))}

如果您的第一列具有聚集但未排序的相等项

1 2	steps_at = np.where(np.diff(np.r_[np.nan, a[:,0]]))[0] return {k:v for k,v in zip(a[steps_at,0], np.split(a[:,1], steps_at[1:]))}

。

一般情况

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ind = np.argsort(a[:, 0], kind='mergesort')
aa = a[ind, 0]
steps_at = np.where(np.diff(np.r_[np.nan, aa]))[0]
return {k:v for k,v in zip(aa[steps_at], np.split(a[ind,1], steps_at[1:]))}

射击：

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(19, 2) correctness

Psidom {0: [0, 28, 38, 97, 99, 65, 73], 1: [64, 91, 70, 40, 9], 2: [94, 96, 69, 46], 3: [85, 15, 65]}
Daniel_Jimenez defaultdict(<class 'list'>, {0: [0, 28, 38, 97, 99, 65, 73], 1: [64, 91, 70, 40, 9], 2: [94, 96, 69, 46], 3: [85, 15, 65]})
Jean_Francois_Fabre {0: [0, 28, 38, 97, 99, 65, 73], 1: [64, 91, 70, 40, 9], 2: [94, 96, 69, 46], 3: [85, 15, 65]}
Alexandre_Kempf {0: array([ 0, 28, 38, 97, 99, 65, 73]), 1: array([64, 91, 70, 40, 9]), 2: array([94, 96, 69, 46]), 3: array([85, 15, 65])}
Or_Duan {0: [0, 28, 38, 97, 99, 65, 73], 1: [64, 91, 70, 40, 9], 2: [94, 96, 69, 46], 3: [85, 15, 65]}
Paul_Panzer_sorted {0: array([ 0, 28, 38, 97, 99, 65, 73]), 1: array([64, 91, 70, 40, 9]), 2: array([94, 96, 69, 46]), 3: array([85, 15, 65])}
Paul_Panzer_grouped {0: array([ 0, 28, 38, 97, 99, 65, 73]), 1: array([64, 91, 70, 40, 9]), 2: array([94, 96, 69, 46]), 3: array([85, 15, 65])}
Paul_Panzer_general {0: array([ 0, 28, 38, 97, 99, 65, 73]), 1: array([64, 91, 70, 40, 9]), 2: array([94, 96, 69, 46]), 3: array([85, 15, 65])}
B_M_sorted {0: array([ 0, 28, 38, 97, 99, 65, 73]), 1: array([64, 91, 70, 40, 9]), 2: array([94, 96, 69, 46]), 3: array([85, 15, 65])}
B_M_general {0: array([ 0, 28, 38, 65, 73, 97, 99]), 1: array([ 9, 40, 64, 70, 91]), 2: array([46, 69, 94, 96]), 3: array([15, 65, 85])}

(40194, 2) speed (seconds used for 10 repeats)

Psidom 0.4336233548820019
Daniel_Jimenez 0.3609276609495282
Jean_Francois_Fabre 0.17962428089231253
Alexandre_Kempf 3.5392782238777727
Or_Duan 0.1873011060524732
Paul_Panzer_sorted 0.08001555898226798
Paul_Panzer_grouped 0.08144942414946854
Paul_Panzer_general 0.10183193604461849
B_M_sorted 0.09192353091202676
B_M_general 0.16612185980193317

(400771, 2) speed (seconds used for 10 repeats)

Psidom 3.968917251098901
Daniel_Jimenez 3.619185874937102
Jean_Francois_Fabre 1.7871235068887472
Or_Duan 1.9176530800759792
Paul_Panzer_sorted 0.8291062880307436
Paul_Panzer_grouped 0.8662846579682082
Paul_Panzer_general 1.0812653130851686
B_M_sorted 1.031000167131424
B_M_general 2.16174431797117
Alexandre_Kempf 513.2718367418274

。

代码：

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from collections import defaultdict
from itertools import groupby
import numpy as np
import timeit

Psidom = lambda a: {k: [v for _, v in g] for k, g in groupby(a, lambda x: x[0])}

def Daniel_Jimenez(a):
d = defaultdict(list)
for x,y in a:
d[x].append(y)
return d

Jean_Francois_Fabre = lambda a: {k:[i[1] for i in v] for k,v in groupby(a,lambda x : x[0])}

def Alexandre_Kempf(a):
keys = a[:,0]
items = a[:,1]
uniqkey = np.unique(keys)
prelist = [items[keys==i] for i in uniqkey]
dico = {}
for i in np.arange(len(uniqkey)):
dico[uniqkey[i]] = prelist[i]
return dico

def Or_Duan(a):
default = {}
for elm in a:
try:
default[elm[0]].append(elm[1])
except KeyError:
default[elm[0]] = [elm[1]]
return default

def Paul_Panzer_sorted(a):
steps_at = np.searchsorted(a[:,0], np.arange(a[-1,0]+1))
return {k:v for k,v in zip(a[steps_at,0], np.split(a[:,1], steps_at[1:]))}

def Paul_Panzer_grouped(a):
steps_at = np.where(np.diff(np.r_[np.nan, a[:,0]]))[0]
return {k:v for k,v in zip(a[steps_at,0], np.split(a[:,1], steps_at[1:]))}

def Paul_Panzer_general(a):
ind = np.argsort(a[:, 0], kind='mergesort')
aa = a[ind, 0]
steps_at = np.where(np.diff(np.r_[np.nan, aa]))[0]
return {k:v for k,v in zip(aa[steps_at], np.split(a[ind,1], steps_at[1:]))}

def B_M_sorted(b):
keys,values=b.T
uniq,steps=np.unique(keys,return_index =True)
bins=np.split(values,steps[1:])
return dict(zip(uniq,bins))

def B_M_general(a):
b=a[np.lexsort(a.T[::-1])]
keys,values=b.T
uniq,steps=np.unique(keys,return_index =True)
bins=np.split(values,steps[1:])
return dict(zip(uniq,bins))

c = np.arange(4).repeat(np.random.randint(1,10,(4)))
d = np.random.randint(100, size=c.shape)
t = np.c_[c, d]
c = np.arange(8000).repeat(np.random.randint(1,10,(8000)))
d = np.random.randint(100, size=c.shape)
a = np.c_[c, d]
c = np.arange(80000).repeat(np.random.randint(1,10,(80000)))
d = np.random.randint(100, size=c.shape)
b = np.c_[c, d]

print(t.shape, 'correctness
')
i = 0
for f in (Psidom, Daniel_Jimenez, Jean_Francois_Fabre, Alexandre_Kempf,
Or_Duan, Paul_Panzer_sorted, Paul_Panzer_grouped,
Paul_Panzer_general, B_M_sorted, B_M_general):
name = f.__name__
if name == '<lambda>':
name = ['Psidom', 'Jean_Francois_Fabre'][i]
i += 1
print(name + (20 - len(name)) * ' ', f(t))

print('
', a.shape, 'speed (seconds used for 10 repeats)
')
i = 0
for f in (Psidom, Daniel_Jimenez, Jean_Francois_Fabre, Alexandre_Kempf,
Or_Duan, Paul_Panzer_sorted, Paul_Panzer_grouped,
Paul_Panzer_general, B_M_sorted, B_M_general):
name = f.__name__
if name == '<lambda>':
name = ['Psidom', 'Jean_Francois_Fabre'][i]
i += 1
print(name + (20 - len(name)) * ' ', timeit.timeit("f(a)",number=10,
globals={'f':f, 'a':a}))

print('
', b.shape, 'speed (seconds used for 10 repeats)
')
i = 0
for f in (Psidom, Daniel_Jimenez, Jean_Francois_Fabre,
Or_Duan, Paul_Panzer_sorted, Paul_Panzer_grouped,
Paul_Panzer_general, B_M_sorted, B_M_general, Alexandre_Kempf):
name = f.__name__
if name == '<lambda>':
name = ['Psidom', 'Jean_Francois_Fabre'][i]
i += 1
print(name + (20 - len(name)) * ' ', timeit.timeit("f(a)",number=10,
globals={'f':f, 'a':b}))

号

相关讨论

很好的一行程序：

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import itertools

array = [[ 0, 1],
[ 1, 0],
[ 1, 2],
[ 2, 3],
[ 2, 1]]

d = {k:[i[1] for i in v] for k,v in itertools.groupby(sorted(array),lambda x : x[0])}

结果：

1	{0: [1], 1: [0, 2], 2: [1, 3]}

。

按列表已排序版本上的第一个项目分组(如果尚未进行排序)
在dict理解中创建字典，只使用分组项的第二个元素构建值列表

相关讨论

假设数组已按第一列排序，则可以使用groupby:

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from itertools import groupby
{k: [v for _, v in g] for k, g in groupby(arr, lambda x: x[0])}
# {0: [1], 1: [0, 2], 2: [3, 1]}

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#arr = np.array([[ 0, 1],
# [ 1, 0],
# [ 1, 2],
# [ 2, 3],
# [ 2, 1]])

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与defaultdict相比，除了使用"请求原谅而不是许可"的方法外，还可以使用"尝试"。

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array = [[ 0, 1],
[ 1, 0],
[ 1, 2],
[ 2, 3],
[ 2, 1]]
default = {}
for elm in array:
try:
default[elm[0]].append(elm[1])
except KeyError:
default[elm[0]] = [elm[1]]

。

相关讨论

一种解决方案是这样做(如果a是您的数组)：

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keys = a[:,0]
items = a[:,1]
uniqkey = np.unique(keys)
prelist = [items[keys==i] for i in uniqkey]
dico = {}
for i in np.arange(len(uniqkey)):
dico[uniqkey[i]] = prelist[i]

。