如果我有这份清单
layout_constraintBottom_toBottomOf
我想将myVar设置为包含数字5的列表的索引。最后myVar将等于1.我该怎么做?我是python的新手,所以一个简单的答案将不胜感激。
答案 0 :(得分:3)
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
答案 1 :(得分:1)
使用for
- 循环从myList
获取子列表,然后检查if 5 in sublist:
这样您就会找到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5
的所有行,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)
答案 2 :(得分:1)
您可以使用enumerate()
创建自定义函数,以获取索引:
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))
示例运行:
>>> 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]]