如何在python中查找包含int的列表的索引

Question

如果我有这个清单清单

myList = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]

并且我想将myVar设置为等于包含数字5的列表的索引。最后，myVar等于1。我该怎么做？ 我是python的新手，因此很高兴获得一个简单的答案。

Answer 1

myList = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
# loop if 5 in the sub_list, than give the index
myVar = [myList.index(ls) for ls in myList if 5 in ls]
print myVar 
# 1

Answer 2

使用for -loop从中获取子列表myList ，然后检查if 5 in sublist:

这样你找到第一行有5

my_list = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
my_var = None

for number, row in enumerate(my_list):
    if 5 in row:
        my_var = number
        break # don't search in other rows

print(my_var)

如果您需要所有带有5行，则列出结果

my_list = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]
my_var = [] # list for all results

for number, row in enumerate(my_list):
    if 5 in row:
        my_var.append( number )

print(my_var)

或一行

my_list = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]

my_var = [number for number, row in enumerate(my_list) if 5 in row]

print(my_var)

Answer 3

您可以使用enumerate()创建自定义函数，以获取索引为：

def get_index(my_list, val):
    for i, item in enumerate(my_list):
        if val in item:
            return i
    else:
        raise ValueError('{} not found in {}'.format(val, my_list))

样品运行：

>>> myList = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]

# Valid value
>>> get_index(myList, 5)
1

# Invalid value
>>> get_index(myList, 22)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 6, in get_index
ValueError: 22 not found in [[1, 2, 3], [4, 5, 6], [7, 8, 9]]