Find values in array and change them efficiently

Question

I am working with a large array with around 300 000 values. It has 100 000 rows and 3 columns. 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. Is there any more efficient way than running something like this?:

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.

Answer 1

Convert your array to a numpy array ( numpy.asarray ) then to replace the values you would use the following:

import numpy as np
N = np.asarray(N)

N[N > 10] = 0

numpy.asarray documentation

Answer 2

I've assumed you may not want to use the same threshold/replacement value for each column. That being the case, you can pack the three items in a list of tuples and iterate through that.

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.

Answer 3

If I understand you correctly you`re looking for something like that:

>>> 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]])