1.zip（）同时迭代多个列表、字典等
使用zip()可以同时迭代多个可迭代对象，如列表、字典。
注意：当若干个可迭代对象的长度不相等时，zip()函数会停止在最短的可迭代对象。
例子：
# 定义可迭代对象
numbers = [1, 2, 3]
letters = ['a', 'b', 'c']
# 使用zip()函数将两个可迭代对象打包，并在循环中同时迭代
for number, letter in zip(numbers, letters):
print(f"Number: {number}, Letter: {letter}")
# 迭代字典
for(triplets, labels), (triplets_h, labels_h) in zip(train_iter, train_iter_h):
if self.p.gpu >= 0:
triplets, labels = triplets.to("cuda"), labels.to("cuda")
triplets_h, labels_h = triplets_h.to("cuda"), labels_h.to("cuda")
subj, rel = triplets[:, 0], triplets[:, 1]
obj, rel = triplets_h[:, 0], triplets_h[:, 1]
2.打包可迭代对象
# 定义可迭代对象
numbers = [1, 2, 3]
letters = ['a', 'b', 'c']
# 使用zip()函数将两个可迭代对象打包
zipped = zip(numbers, letters)
# 将zipped转换为列表并打印
zipped_list = list(zipped)
print(zipped_list) # 输出：[(1, 'a'), (2, 'b'), (3, 'c')]
3.*操作符解包
# 定义可迭代对象
numbers = [1, 2, 3]
letters = ['a', 'b', 'c']
# 使用zip()函数将两个可迭代对象打包，并解包合并为多个列表
zipped = zip(numbers, letters)
merged_numbers, merged_letters = zip(*zipped)
print("Merged numbers:", merged_numbers) # 输出：Merged numbers: (1, 2, 3)
print("Merged letters:", merged_letters) # 输出：Merged letters: ('a', 'b', 'c')
