根据每个三元组的第二个元素在三元组列表中进行分层

问题描述

在python中，我有以下形式的三元组列表：

[(220,1.0,1),(385,2),(405,(1276,6),(1649,8),(1941,10),(2554,13),(3123,16),(2377,0.8879465659,12),(500,0.8854919047,(2435,0.8815715038,(2151,0.8787807797,11),(1888,0.87827976,9),(2185,0.8780501222,(2215,0.8747450062,(358,0.8724861947,(3636,0.8716343914,19),(734,0.8714647102,3),(1742,0.8707242976,…………]

我想将列表中的元素按照列表中每个三元组的第二个值，分成十个轨道：

=0 和

=0.1 和

=0.2 和

=0.3 和

= 0.4 和

.....

=0.8 和

=0.9 和

我尝试使用：

tracks =  np.linspace(0,1,11)  #  array([0.,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1. ])
for i in range (len(tracks)-1):
    print( tracks[i],"a",tracks[i+1],":",sum([1 for (x,y,z) in sm_list   if y >= tracks[i]] and y < intervalos[i+1]))

但它返回以下错误：“NameError: name 'y' is not defined”

如何定义复合布尔子句以形成上述列表？

制作：

tracks =  np.linspace(0,z) in sm_list   if y >= tracks[i]] ))

不返回错误，但返回重叠的轨道，这不是我需要的。

解决方法

• 首先要注意：您在第一个示例中有错别字。以下内容基本上可以满足您的需求：

tracks =  np.linspace(0,1,11)  #  array([0.,0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,1. ])
for i in range (len(tracks)-1):
    print( tracks[i],"a",tracks[i+1],":",sum([1 for (x,y,z) in sm_list   if y >= tracks[i] and y < tracks[i+1]]))

• 为什么是“本质上”？因为最右边的边界，即 tracks[-1]，等于 1.0，所以 y < 1.0 不会捕获那些是 ==1 的值。快速且脏修复：在进入循环之前设置 tracks[-1] = 1.01。

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• 奖励：原始问题的另一种解决方案，它不使用 for 循环，但更喜欢 numpy 方法，带有一点 pandas :（我鼓励您运行在您的数据上逐行查看输出；也许您会意识到您只需要前几行，这取决于您的真正目的）

numbers = [y for (x,z) in sm_list]  # we care only about the second number in each triple anyway
x = (np.array(numbers) * 10).astype(np.int)  # multiply by 10 and "floor" into ints in the range 0,2,...,10 (e.g. 0-0.9999 will be mapped to 0,1.0-1.9999 to 1,9.0-9.9999 to 9,and 1.0 to 10)

# Now we have a list of integers between 0 and 10,and we remain with the task of getting their histogram or counts.

# There are plenty of ways to solve this,e.g. using collections.Counter(x),but we'll use a fancier way:
s = pd.value_counts(x)  # already may be suffice for your needs
s = s.reindex(range(0,11),fill_value=0)  # try this to ensure that all values 0,10 have some count associated with them (namely 0) even if they did not exist in the input

if False:  # bonus: if you don't like the cell s[10] and want its value to be added into s[9],then change this 'if False' to 'if True'...
    s[9] += s[10];
    s.drop(10,inplace=True)