微信公众号搜"智元新知"关注
微信扫一扫可直接关注哦!

python – 有效地按元素分组

可以说我有

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

list = [[500,800,1000,200,1500],[220,450,350,1070,1780],[900,1780,1450,100],[340,670,830,1370,1420],[850,630,1230,1670,910]]

angle = [[50,80,100,20,150],[22,45,35,107,178],[90,178,145,10],[34,67,83,137,142],[85,63,123,167,91]]

我想将每个元素放在列表中,并根据其值将其存储在不同的单独数组中;

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 = [500,340] 
for all list.values where angles.value is between 60 and 90
list3 = [800,850,630]

等等..

我做了这样的事情:

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])

我想知道是否有一种pythonic /有效的方法来提高性能.

解决方法

这是一个矢量化的方法

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,np.inf])
    ...: 
    ...: list1 = np.array([[500,\
    ...:                   [220,\
    ...:                   [900,...:                   [340,\
    ...:                   [850,910]])
    ...: 
    ...: angle = np.array([[50,\
    ...:                   [22,\
    ...:                   [90,...:                   [34,\
    ...:                   [85,91]])
    ...: 

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

版权声明:本文内容由互联网用户自发贡献,该文观点与技术仅代表作者本人。本站仅提供信息存储空间服务,不拥有所有权,不承担相关法律责任。如发现本站有涉嫌侵权/违法违规的内容, 请发送邮件至 dio@foxmail.com 举报,一经查实,本站将立刻删除。

相关推荐