列表：查找第一个索引并计算列表列表中特定列表的出现次数

Question

我们有一个名为 location 的变量。

location=[["world", 'Live'], ["alpha",'Live'], ['hello', 'Scheduled'],['alpha', 'Live'], ['just', 'Live'], ['alpha','Scheduled'], ['alpha', 'Live']]

我想在位置找到第一个索引并计算list["alpha",'Live']的出现次数。 我尝试了以下方法：

index= [location.index(i) for i in location if i ==["alpha", 'Live'] ]
count = [location.count(i) for i in location if i ==["alpha",'Live'] ]
print('index',index)
print('count', count)

这将返回： index [1, 1, 1] count [3, 3, 3]

但是有没有办法找到第一个索引，使用列表理解同时计数。

预期 output：

索引，计数 = 1, 3

Answer 1

这能解决你的问题吗？

location=[["world", 'Live'], ["alpha",'Live'], ['hello', 'Scheduled'],['alpha', 'Live'], ['just', 'Live'], ['alpha','Scheduled'], ['alpha', 'Live']]
index= location.index(["alpha",'Live'])
count = location.count(["alpha",'Live'])
print('index',index)
print('count', count)

如果未找到['alpha','live'] ，则找到第一个['alpha',??]并打印其索引和计数。

location = [["world", 'Live'], ["alpha", 'Live'], ['hello', 'Scheduled'], [
    'alpha', 'Live'], ['just', 'Live'], ['alpha', 'Scheduled'], ['alpha', 'Live']]

key = ["alpha", 'Lsive']

count = location.count(key)

if count:
    index = location.index(key)
    print('count', count)
    print('index', index)
else:
    for i in location:
        if i[0] == key[0]:
            key = i
            count = location.count(key)
            index = location.index(key)
            print('count', count)
            print('index', index)
    else:
        print('not found')

@yadavender yadav 编写的更简洁的代码

location = [["alpha", 'Scheduled'], ["alpha", 'Live'], ['hello', 'Scheduled'], [
    'alpha', 'Live'], ['just', 'Live'], ['alpha', 'Scheduled'], ['alpha', 'Live']]

key = ["alpha", 'Scheduled']

count = location.count(key)

if count:
    index = location.index(key)
else:
    index=[location.index(i) for i in location if i[0]=="alpha"][0]
print('count', count)
print('index', index)