[英]Python: Counting Zeros in multiple array columns and store them efficently
I create an array:我创建一个数组:
import numpy as np
arr = [[0, 2, 3], [0, 1, 0], [0, 0, 1]]
arr = np.array(arr)
Now I count every zero per column and store it in a variable:现在我计算每列的每个零并将其存储在一个变量中:
a = np.count_nonzero(arr[:,0]==0)
b = np.count_nonzero(arr[:,1]==0)
c = np.count_nonzero(arr[:,2]==0)
This code works fine.此代码工作正常。 But in my case I have many more columns with over 70000 values in each.
但在我的情况下,我有更多的列,每个列都有超过 70000 个值。 This would be many more lines of code and a very messy variable expolorer in spyder.
这将是更多的代码行和 spyder 中一个非常混乱的变量探索器。
My questions:我的问题:
Thank you谢谢
You can construct a boolean array arr == 0
and then take its sum along the rows.您可以构造一个 boolean 数组
arr == 0
,然后沿行取其总和。
>>> (arr == 0).sum(0)
array([3, 1, 1])
Use an ordered dict from the collections module:使用 collections 模块中的有序字典:
from collections import OrderedDict
import numpy as np
from pprint import pprint as pp
import string
arr = np.array([[0, 2, 3], [0, 1, 0], [0, 0, 1]])
letters = string.ascii_letters
od = OrderedDict()
for i in range(len(arr)):
od[letters[i]] = np.count_nonzero(arr[:, i]==0)
pp(od)
Returning:返回:
OrderedDict([('a', 3), ('b', 1), ('c', 1)])
Example usage:示例用法:
print(f"First number of zeros: {od.get('a')}")
Will give you:会给你:
First number of zeros: 3
To count zeros you can count non-zeros along each column and subtract result from length of each column:要计算零,您可以沿每列计算非零并从每列的长度中减去结果:
arr.shape[0] - np.count_nonzero(arr, axis=0)
produces [3,1,1]
.产生
[3,1,1]
。
This solution is very fast because no extra large objects are created.这个解决方案非常快,因为没有创建额外的大对象。
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