如何解决根据每个三元组的第二个元素在三元组列表中进行分层
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
。
- 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)
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。