[英]Numpy mask array multiple times and fill nans in 3D array with values from another 3D array
我有以下代码:
import numpy as np
def fill(arr1, arr2, arr3, arr4, thresh= 0.5):
out_arr = np.zeros(arr1.shape)
for i in range(0,len(arr1)):
arr1[i] = np.where(np.abs(arr1[i])<=thresh,np.nan,arr1[i])
mask = np.isnan(arr1[i])
arr1[i] = np.nan_to_num(arr1[i])
merged1 = (arr2[i]*mask)+arr1[i]
merged2 = np.where(np.abs(merged1)<=thresh,np.nan,merged1)
mask = np.isnan(merged2)
merged2 = np.nan_to_num(merged2)
merged3 = (arr3[i]*mask)+merged2
merged3 = np.where(np.abs(merged3)<=thresh,np.nan,merged3)
mask = np.isnan(merged3)
merged3 = np.nan_to_num(merged3)
merged4 = (arr4[i]*mask)+merged3
out_arr[i] = merged4
return(out_arr)
arr1 = np.random.rand(10, 10, 10)
arr2 = np.random.rand(10, 10, 10)
arr3 = np.random.rand(10, 10, 10)
arr4 = np.random.rand(10, 10, 10)
arr = fill(arr1, arr2, arr3, arr4, 0.5)
我想知道是否有更有效的方法可以使用屏蔽的 arrays? 基本上我所做的是将 3D 阵列的每一层中低于阈值的值替换为下一个阵列,这超过 4 个 arrays。 对于 n arrays,这会是什么样子? 谢谢!
您的 function 可以通过多种方式进行简化。 在效率方面,最重要的方面是您不需要迭代第一维,您可以直接对整个 arrays 进行操作。 除此之外,您可以将替换逻辑重构为更简单的东西,并使用循环来避免一遍又一遍地重复相同的代码:
import numpy as np
# Function accepts as many arrays as wanted, with at least one
# (threshold needs to be passed as keyword parameter)
def fill(arr1, *arrs, thresh=0.5):
# Output array
out_arr = arr1.copy()
for arr in arrs:
# Replace values that are still below threshold
mask = np.abs(out_arr) <= thresh
out_arr[mask] = arr[mask]
return out_arr
由于thresh
需要在此 function 中作为关键字参数传递,因此您可以将其称为:
arr = fill(arr1, arr2, arr3, arr4, thresh=0.5)
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