Python 按元素对数组进行有效分组

假设我有

lags = [0, 30, 60, 90, 120, 150, 180, np.inf]

及

我希望获得列表中的每个元素，并根据其值将其存储在不同的单独数组中，以便：

for all list.values where angles.value is less than 30
list1 = [200, 220, 100]
for all list.values where angles.value is between 30 and 60
list2 = [500, 450, 350, 450, 340] 
for all list.values where angles.value is between 60 and 90
list3 = [800, 670, 830, 850, 630]

等等

我是这样做的：

sortlist = defaultdict(list)
ulist = np.unique(list)
uangle = np.unique(angle)
for lag in lags:
    count += 1
    for k, dummy_val in enumerate(uangle):
        if lag <= uangle[k] < lag + 1:
            sortlist[count].append(ulist[k])

sortlist=defaultdict（列表）
ulist=np.unique（列表）
uangle=np.唯一（角度）
对于滞后中的滞后：
计数+=1
对于k，枚举中的伪值（uangle）：
如果使用numpy滞后：
angle这是一种矢量化方法-

an = angle.ravel()
sidx = an.argsort()
cut_idx = np.searchsorted(an[sidx], lags)
out = np.split(list1.ravel()[sidx], cut_idx[1:-1])

样本输入、输出-

In [97]: lags = np.array([0, 30, 60, 90, 120, 150, 180, np.inf])
    ...: 
    ...: list1 = np.array([[500, 800, 1000, 200, 1500], \
    ...:                   [220, 450, 350, 1070, 1780], \
    ...:                   [900, 450, 1780, 1450, 100], 
    ...:                   [340, 670, 830, 1370, 1420], \
    ...:                   [850, 630, 1230, 1670, 910]])
    ...: 
    ...: angle = np.array([[50, 80, 100, 20, 150],\
    ...:                   [22, 45, 35, 107, 178],\
    ...:                   [90, 45, 178, 145, 10], 
    ...:                   [34, 67, 83, 137, 142],\
    ...:                   [85, 63, 123, 167, 91]])
    ...: 

In [99]: out
Out[99]: 
[array([100, 200, 220]),            # <----- 0 to 30
 array([340, 350, 450, 450, 500]),  # <----- 30 to 60
 array([630, 670, 800, 830, 850]),  # <----- 60 to 90
 array([ 900,  910, 1000, 1070]),   # <----- 90 to 120
 array([1230, 1370, 1420, 1450]),   # <----- 120 to 150
 array([1500, 1670, 1780, 1780]),   # <----- 150 to 180
 array([], dtype=int64)]            # <----- 180 to Inf

[97]中的
lags=np.array（[0,30,60,90,120,150,180，np.inf]）
...: 
…：list1=np.array（[50080010002001500]\
...:                   [220, 450, 350, 1070, 1780], \
...:                   [900, 450, 1780, 1450, 100], 
...:                   [340, 670, 830, 1370, 1420], \
...:                   [850, 630, 1230, 1670, 910]])
...: 
…：角度=np.数组（[50,80,100,20,150]\
...:                   [22, 45, 35, 107, 178],\
...:                   [90, 45, 178, 145, 10], 
...:                   [34, 67, 83, 137, 142],\
...:                   [85, 63, 123, 167, 91]])
...: 
In[99]：out
出[99]：
[array（[100200220]），#解决方案，基于在普通Python中实现它（不含numpy）

您可以将值存储在
dict
中，而不是存储在单独的变量中（使用是更好的选择）

您可以创建一个函数，根据角度返回组：

def get_group_from_angle(angle):
    group = ''
    if angle < 30:
        group = 'a'
    elif 30 < angle < 60:
        group = 'b' 
    elif 60 < angle < 90:
        group = 'c'
    return group

my_dict
保存的最终值为：

{ 
    'a': [200, 220, 100], 
    'b': [500, 450, 350, 450, 340],
    'c': [800, 670, 830, 850, 630], 
    '': [1000, 1500, 1070, 1780, 900, 1780, 1450, 1370, 1420, 1230, 1670, 910]
    # ^ number whose angle is not present in any specified range 
}

您关心每个输出列表中元素的顺序吗？@Divakar:我不关心顺序。@Divakar，只是删除了它。如果需要，我必须添加“=”。@user7436576更新了滞后中的所有值。为什么
剪切_idx[1:-1]
？@MYGz Well
np.split
只需要间隔重叠的中间索引就可以拆分为数组。因此，我们不需要起始索引和结束索引。
[200, 220, 100]
[500, 450, 350, 450, 340]
[800, 670, 830, 850, 630]
[1000, 1070, 900, 910]
[1450, 1370, 1420, 1230]
[1500, 1780, 1780, 1670]
[]

an = angle.ravel()
sidx = an.argsort()
cut_idx = np.searchsorted(an[sidx], lags)
out = np.split(list1.ravel()[sidx], cut_idx[1:-1])

In [97]: lags = np.array([0, 30, 60, 90, 120, 150, 180, np.inf])
    ...: 
    ...: list1 = np.array([[500, 800, 1000, 200, 1500], \
    ...:                   [220, 450, 350, 1070, 1780], \
    ...:                   [900, 450, 1780, 1450, 100], 
    ...:                   [340, 670, 830, 1370, 1420], \
    ...:                   [850, 630, 1230, 1670, 910]])
    ...: 
    ...: angle = np.array([[50, 80, 100, 20, 150],\
    ...:                   [22, 45, 35, 107, 178],\
    ...:                   [90, 45, 178, 145, 10], 
    ...:                   [34, 67, 83, 137, 142],\
    ...:                   [85, 63, 123, 167, 91]])
    ...: 

In [99]: out
Out[99]: 
[array([100, 200, 220]),            # <----- 0 to 30
 array([340, 350, 450, 450, 500]),  # <----- 30 to 60
 array([630, 670, 800, 830, 850]),  # <----- 60 to 90
 array([ 900,  910, 1000, 1070]),   # <----- 90 to 120
 array([1230, 1370, 1420, 1450]),   # <----- 120 to 150
 array([1500, 1670, 1780, 1780]),   # <----- 150 to 180
 array([], dtype=int64)]            # <----- 180 to Inf

def get_group_from_angle(angle):
    group = ''
    if angle < 30:
        group = 'a'
    elif 30 < angle < 60:
        group = 'b' 
    elif 60 < angle < 90:
        group = 'c'
    return group

from collections import defaultdict
my_dict = defaultdict(list)

#  `alist` and `angle` are variables holding values as mentioned in Question

for ll, aa in zip(list, angle):
    for l, a in zip(ll, aa):
        my_dict[get_group_from_angle(a)].append(l)

{ 
    'a': [200, 220, 100], 
    'b': [500, 450, 350, 450, 340],
    'c': [800, 670, 830, 850, 630], 
    '': [1000, 1500, 1070, 1780, 900, 1780, 1450, 1370, 1420, 1230, 1670, 910]
    # ^ number whose angle is not present in any specified range 
}