从for循环

Question

好的，所以我有两个列表，它们可以是从一个值长到20的任何位置，但它们总是具有相同的数量。

例如

alphabet = ['a', 'b', 'c', 'd', 'e']
numbers = ['1', '2', '3', '4', '5']

现在我的目标是创建一个遍历两个列表的for循环，并将每个列表中的相应值相互添加。所以..

['a1', 'b2', 'c3', 'd4', '5e']

再举一个例子。

names = ['john', 'harry', 'joe']
IDs = ['100', '200', '300']

output: ['john100', 'harry200', 'joe300']

有人能指出我正确的方向吗？

Answer 1

您可以使用from selenium import webdriver driver = webdriver.Firefox() driver.get('http://www.website.com') source = driver.page_source location = source.find('<p class="info"') source = source[location+16:] #adjust the +16 to maybe +15 or w/e depending on the exact source page you get location_second = source.find('viewers on') #assuming there is a space between the actual number of viewers and the source = int(source[:location_second-1]) #adjust the -1 to maybe -2 or w/e depending on the exact source page you get if source > x: # replace x with whatever number is your minimum viewers driver.find_element_by_class_name('js-profile-link') #might need to use x-path if you have multiple instances of the same class name 和zip：

join

至于第二个例子：

[''.join(p) for p in zip(alphabet, numbers)]
# ['a1', 'b2', 'c3', 'd4', 'e5']

Answer 2

作为@Psidom单线程解决方案的替代方案，您可以使用zip()：

>>>[i+j for i, j in zip(alphabet, numbers)]
>>>['a1', 'b2', 'c3', 'd4', 'e5']

或者如果你完全使用常规for循环：

res = []
for i, j in zip(alphabet, numbers):
    res.append(i+j)

你也可以把它变得更通用，并把它放在一个函数中：

# method one
def concancate_elements(list1, list2):
    return [i+j, for i, j in zip(alphabet, numbers)]

# method two
def concancate_elements(list1, list2):
    res = []
    for i, j in zip(alphabet, numbers):
        res.append(i+j)
    return res