打印 numpy 数组中所有列都满足特定条件的行？

Question

I have an array with 500 rows and 5 columns. I need to find all the rows where the value in each of the last 4 columns is greater than 100. I found a way to check each column individually but I'd like to be able to check them all at once. I tried inserting an axis argument but it gives me an error. There must be a simpler way to do this. This is what I could get to work:

over1 = (array[:,1] >= 100)
over2 = (array[:,2] >= 100)
over3 = (array[:,3] >= 100)
over4 = (array[:,4] >= 100)
where = np.argwhere(over1&over2&over3&over4 == True)
there = array[where]
there2 = np.array(there[:,0]) 
#I had to reshape because the new array was a different shape for some reason

I'm new to Python and Numpy so I'm having some trouble

Answer 1

I believe you're looking for:

x[(x[:, 1:] > 100).all(axis=1)]

---

Consider x :

print(x)
array([[ 79, 192, 163,  94, 186],
       [111, 183, 152, 115, 171],
       [ 61, 125,  91, 163,  60],
       [110,  24,   0, 151, 180],
       [165, 111, 141,  19,  81]])

The operation x[:, 1:] > 100 broadcasts the operation on every element, resulting in a boolean matrix.

print(x[:, 1:] > 100)
array([[ True,  True, False,  True],
       [ True,  True,  True,  True],
       [ True, False,  True, False],
       [False, False,  True,  True],
       [ True,  True, False, False]], dtype=bool)

np.all , similar to the inbuilt function all , will evaluate to True if every element is True , else it evaluates to False . We want to perform this check every column per row, so we need axis=1 .

mask = (x[:, 1:] > 100).all(1)
print(mask)
Out[362]: array([False,  True, False, False, False], dtype=bool)

The mask will now be used to index into the original.

x[mask]
array([[111, 183, 152, 115, 171]])