["python","numpy","nan"]
convert nan value to zero

I have a 2D numpy array. Some of the values in this array are NaN. I want to perform certain operations using this array. For example consider the array: [[ 0. 43. 67. 0. 38.] [ 100. 86. 96. 100. 94.] [ 76. 79. 83. 89. 56.] [ 88. NaN 67. 89. 81.] [ 94. 79. 67. 89. 69.] [ 88. 79. 58. 72. 63.] [ 76. 79. 71. 67. 56.] [ 71. 71. NaN 56. 100.]] I am trying to take each row, one at a time, sort it in reversed order to get max 3 values from the row and take their average. The code I tried is: # nparr is a 2D numpy array for entry in nparr: sortedentry = sorted(entry, reverse=True) highest_3_values = sortedentry[:3] avg_highest_3 = float(sum(highest_3_values)) / 3 This does not work for rows containing NaN. My question is, is there a quick way to convert all NaN values to zero in the 2D numpy array so that I have no problems with sorting and other things I am trying to do.
This should work: from numpy import * a = array([[1, 2, 3], [0, 3, NaN]]) where_are_NaNs = isnan(a) a[where_are_NaNs] = 0 In the above case where_are_NaNs is: In [12]: where_are_NaNs Out[12]: array([[False, False, False], [False, False, True]], dtype=bool) A complement about efficiency. The examples below were run with numpy 1.21.2 >>> aa = np.random.random(1_000_000) >>> a = np.where(aa < 0.15, np.nan, aa) >>> %timeit a[np.isnan(a)] = 0 536 µs ± 8.11 µs per loop (mean ± std. dev. of 7 runs, 1,000 loops each) >>> a = np.where(aa < 0.15, np.nan, aa) >>> %timeit np.where(np.isnan(a), 0, a) 2.38 ms ± 27.8 µs per loop (mean ± std. dev. of 7 runs, 100 loops each) >>> a = np.where(aa < 0.15, np.nan, aa) >>> %timeit np.nan_to_num(a, copy=True) 8.11 ms ± 401 µs per loop (mean ± std. dev. of 7 runs, 100 loops each) >>> a = np.where(aa < 0.15, np.nan, aa) >>> %timeit np.nan_to_num(a, copy=False) 3.8 ms ± 70.2 µs per loop (mean ± std. dev. of 7 runs, 100 loops each) In consequence a[np.isnan(a)] = 0 is faster.