I have a large 2D array, array, with each entry being a large array of numbers:
array = [
[1, 0, 3, ...],
[5, 4, 1, ...],
[1, 2, 3, ...],
...
]
All the numbers in the 2D array are from 0-5 and I have to somehow find and replace specific numbers, like, for example, all occurrences of the number 3 and replace it with 5.
This can be done with list comprehension in a simple one-liner.
Say you have a list of lists:
a = [[1, 2, 3], [1, 2, 3], [1, 2, 3]]
and you want to replace all occurrences of 2 with 4:
[[_el if _el != 2 else 4 for _el in _ar] for _ar in a]
Another option is to use numpy's where function. From the docstring:
where(condition, [x, y])
Return elements, either from
xory, depending oncondition.
So, in your case (say you'd like again to replace all 2 with 4):
import numpy as np
a = np.array([[1, 2, 3],
[1, 2, 3],
[1, 2, 3]])
np.where(a==2, 4, a)
If you want to replace several values in one go, you could do something like this:
Say you'd like to replace 1 with 3 and 3 with 5:
ix=np.isin(array, [1,3])
vc=np.vectorize(lambda x: 3 if x == 1 else 5)
np.where(ix, vc(array), array)
If you have more than 2 values to replace, say you want to map the list [1,3,5] to [3, 5, -3], then you can define a simple function like:
old_vals = [1,3,5]
new_vals = [3, 5, -3]
def switch_val(x):
return new_vals[old_vals.index(x)] if x in old_vals else x
and so:
vc=np.vectorize(switch_val)
vc(array)
where we vectorized the function.
Hope that helped and happy coding!
for i in range(len(array)):
for j in range(len(array[i])):
if array[i][j] == value_you_are_looking_for:
array[i][j] = new_value
This should work
For a single value, given you have an array called Arr, such that:
Arr = np.array([[1,2,3],
[1,2,3],
[1,2,3]])
Then:
mask = Arr == 2
gives:
mask = np.array([[False,True,False],
[False,True,False],
[False,True,False]])
finally, replacing 2 with 4:
newArr[mask] = 4
gives:
Arr = np.array([[1,4,3],
[1,4,3],
[1,4,3]])