[英]Find values in array and change them efficiently
I am working with a large array with around 300 000 values. 我正在处理大约30万个值的大型数组。 It has 100 000 rows and 3 columns.
它具有10万行和3列。 I am doing iterations with this array and if any value in the first column exceeds a limit of lets say 10, I want the number to be replaced.
我正在使用此数组进行迭代,如果第一列中的任何值超过了可以说10的限制,我希望替换该数字。 Is there any more efficient way than running something like this?:
有没有比运行类似这样更有效的方法?:
for i in range(N):
if array[i][0]>10:
array[i][0] = 0
I need to repeat this sequense for the other two columns as well which included with all my other iterations makes my code pretty slow. 我还需要对其他两列重复此顺序,所有其他迭代中都包含该顺序,这会使我的代码运行缓慢。
Convert your array to a numpy array ( numpy.asarray
) then to replace the values you would use the following: 将您的数组转换为numpy数组(
numpy.asarray
),然后使用以下内容替换值:
import numpy as np
N = np.asarray(N)
N[N > 10] = 0
I've assumed you may not want to use the same threshold/replacement value for each column. 我假设您可能不想为每列使用相同的阈值/替换值。 That being the case, you can pack the three items in a list of tuples and iterate through that.
在这种情况下,您可以将三个项目打包在一个元组列表中并进行遍历。
import numpy as np
arr = np.ndarray(your_array)
#Edited with your values, and a more than symbol
threshold = 10
column_id = 0
replace_value = 0
arr[arr[:, column_id] > threshold, column_id] = replace_value
Set threshold
, column_id
and replace_value
as you require. 根据需要设置
threshold
, column_id
和replace_value
。
If I understand you correctly you`re looking for something like that: 如果我对您的理解正确,那么您正在寻找类似的东西:
>>> from numpy import array
>>> a = array([[1,2,3],[4,5,6],[7,8,9]])
>>> a
array([[1, 2, 3],
[4, 5, 6],
[7, 8, 9]])
>>> a[a>5]=10 # <--- here the "magic" happens
>>> a
array([[ 1, 2, 3],
[ 4, 5, 10],
[10, 10, 10]])
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