Menu
I have an array of shape 512x512 which contains numbers between 0 and 100 at ith and jth position. Now I want to select array[i,j] < 25 and zero at other places. I have tried with array = array[where(array<25)], which gives me a 1D array, but I want 2D. Please help me to solve this.
235
min to return the minimum value, or equivalently for an array arrname use arrname. min() . As you mentioned, numpy. argmin returns the index of the minimum value (of course, you can then use this index to return the minimum value by indexing your array with it).
datetime64() method, we can get the date in a numpy array in a particular format i.e year-month-day by using numpy. datetime64() method. Syntax : numpy.datetime64(date) Return : Return the date in a format 'yyyy-mm-dd'.
NumPy allows the subtraction of two datetime values, an operation which produces a number with a time unit. Because NumPy doesn't have a physical quantities system in its core, the timedelta64 data type was created to complement datetime64 .
One solution:
result = (array < 25) * array
The first part array < 25 gives you an array of the same shape that is 1 (True) where values are less than 25 and 0 (False) otherwise. Element-wise multiplication with the original array retains the values that are smaller than 25 and sets the rest to 0. This does not change the original array
Another possibility is to set all values that are >= 25 to zero in the original array:
array[array >= 25] = 0
I also wanted to add that you can take advantage of numpy views to achieve this:
>>> a = np.asarray([ [1,2], [3,4], [4,1], [6,2], [5,3], [0,4] ])
>>> b = a[:, 1] # lets say you only care about the second column
>>> b[b > 3] = 0
>>> print(a)
[[1 2]
[3 0]
[4 1]
[6 2]
[5 3]
[0 0]]
This is nice when you want the values to be something other than 0.
43
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With
