您将如何比较两个字符串，以便它们可以循环并维护每个标头的项目?提供的示例

Question

我在打印这两个字符串(或只有一个(时遇到问题，因为它们将它们附加到它们的标题并分组。 新标头 (h*( 的指示是新组对开始的地方。

headers = ['h1', 'h2', 'h3']
full = ['h1','a','b','c','h2','d','e','f','h3','g','h','i','j','k']

print('h1' + ':' + 'a')
print('h1' + ':' + 'b')
print('h1' + ':' + 'c')
print('h2' + ':' + 'd')
print('h2' + ':' + 'e')
print('h2' + ':' + 'f')
print('h3' + ':' + 'g')
etc.
output:   
h1:a
h1:b
h1:c
h2:d
etc.

Answer 1

# Initialize index to -1.
header_i = -1
# Go through each element.
for elem in full:
    # If element is in the headers list, update header index to element's index in headers.
    if elem in headers:
        header_i = headers.index(elem)
    # Else, print out the current header and element.
    else:
        print(headers[header_i], ':', elem)