Python：沿轴的矢量化条件和

Question

是否有一种方法可以在某些条件下沿轴求和？所以我有一个像这样的数组，可以沿0轴求和。

#array creation:
tpfnfpArray = np.zeros((1000,3))
for i in range(1000):
    tpfnfpArray[i,:] = (i,i,i)

#first result
tp,fn, fp = np.sum(tpfnfpArray,axis=0) 

#preparing second result
tp2,fp2,fn2,tn2 = (0,0,0,0)

到目前为止很好。现在我想在这样的条件下得到另一个结果（tp2，fp2，fn2，tn2）：

for i in range(1000):
    if tpfnfpArray[i][0] > 0 or tpfnfpArray[i][1]>0:
        if tpfnfpArray[i][2] > 0: # 0,1,1 or 1,0,1
            tp2+=1
        else: # 0,1,0 or 1,0,0
            fp2+=1
    else:
        if tpfnfpArray[i][2] > 0: # 0,0,1
            fn2+=1
        else: # 0,0,0
            tn2+=1

是否有可能无需循环遍历第一个数组的每一行？

Answer 1

您可以使用矢量化布尔运算来进行该计算：

import numpy as np

# Random binary array
np.random.seed(0)
tpfnfpArray = np.random.randint(0, 2, (1000, 3))

# Loop computation for comparison
tp2, fp2, fn2, tn2 = (0, 0, 0, 0)
for i in range(1000):
    if tpfnfpArray[i][0] > 0 or tpfnfpArray[i][1]>0:
        if tpfnfpArray[i][2] > 0:
            tp2 += 1
        else:
            fp2 += 1
    else:
        if tpfnfpArray[i][2] > 0:
            fn2 += 1
        else:
            tn2 += 1
print(tp2, fp2, fn2, tn2)
# 401 377 115 107

# Vectorized computation
tp_m = tpfnfpArray[:, 0] > 0
fn_m = tpfnfpArray[:, 1] > 0
fp_m = tpfnfpArray[:, 2] > 0
tpfn_m = tp_m | fn_m
tp3 = np.count_nonzero(tpfn_m & fp_m)
fp3 = np.count_nonzero(tpfn_m & ~fp_m)
fn3 = np.count_nonzero(~tpfn_m & fp_m)
tn3 = np.count_nonzero(~tpfn_m & ~fp_m)
print(tp3, fp3, fn3, tn3)
# 401 377 115 107