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python 对多个数组的对应位置的元素进行处理(做差再平方)


实例:对多个数组的对应位置的元素进行处理
对图像的相似度进行处理,用到了这一步,所以写了这个
如需要查看 ​​​图像相似度请点击​​

diff_list_all = [[1,2,3],
[4,5,6],
[10,10,10]]

sum_all = []
for i in range(len(diff_list_all)):
for j in range(len(diff_list_all[i])):
for k in range(i+1,(len(diff_list_all))):
diffe_squra = (diff_list_all[i][j] - diff_list_all[k][j]) ** 2
sum_all.append(diffe_squra)
print("sum_all:", sum_all)

sum_All = []
for i in range(len(diff_list_all)):
sum_All.append(sum_all[(len(diff_list_all)-1)*i])
print("sum_All_1:",sum_All)

# for i in range(len(diff_list_all)):
# sum_All.append(sum_all[(len(diff_list_all)-1)*i+1])
# print("sum_All_2:",sum_All)
"""
依次计算 (1-4)**2; (1-10)**2
(2-5)**2; (2-10)**2
(3-6)**2; (3-10)**2
(4-10)**2; (5-10)**2;(6-10)**2
"""
输出:
sum_all: [9, 81, 9, 64, 9, 49, 36, 25, 16]
sum_All_1: [9, 9, 9]

注释部分输出
sum_all: [9, 81, 9, 64, 9, 49, 36, 25, 16]
sum_All_2: [81, 64, 49]

封装函数

diff_list_all = [[1,2,3],
[4,5,6],
[10,10,10]]

def get_sum_all(diff_list_all):
sum_all = []
for i in range(len(diff_list_all)):
for j in range(len(diff_list_all[i])):
for k in range(i + 1, (len(diff_list_all))):
diffe_squra = (diff_list_all[i][j] - diff_list_all[k][j]) ** 2
sum_all.append(diffe_squra)
return sum_all

def double_img_all(n,sum_all,diff_list_all):
sum_All = []
if n>=0 and n<len(diff_list_all):
for i in range(len(diff_list_all)):
sum_All.append(sum_all[(len(diff_list_all)-1)*i+n])
print("sum_All:", sum_All)
return sum_All

sum_all = get_sum_all(diff_list_all)
for n in range(len(diff_list_all)-1): #遍历输出: 数组1与数组2的计算值;数组1与数组3的计算值
double_img_all(n,sum_all,diff_list_all)

输出:
sum_All: [9, 9, 9]
sum_All: [81, 64, 49]

封装成一个函数

diff_list_all = [[1,2,3],
[4,5,6],
[10,10,10]]

def double_img_all(n,diff_list_all):
sum_All = []
sum_all = []
for i in range(len(diff_list_all)):
for j in range(len(diff_list_all[i])):
for k in range(i + 1, (len(diff_list_all))):
diffe_squra = (diff_list_all[i][j] - diff_list_all[k][j]) ** 2
sum_all.append(diffe_squra)
if n>=0 and n<len(diff_list_all):
for i in range(len(diff_list_all)):
sum_All.append(sum_all[(len(diff_list_all)-1)*i+n])
else:
print("error number")
return sum_all,sum_All

for n in range(len(diff_list_all)-1): #遍历输出: 数组1与数组2的计算值;数组1与数组3的计算值
sum_all, sum_All = double_img_all(n,diff_list_all)
print("sum_All:", sum_All)
print("sum_all:", sum_all)

输出:
sum_All: [9, 9, 9]
sum_all: [9, 81, 9, 64, 9, 49, 36, 25, 16]
sum_All: [81, 64, 49]
sum_all: [9, 81, 9, 64, 9, 49, 36, 25, 16]


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