Python 列表中的列表

Question

我有 1 个主列表，它基于 2 个子列表。我想创建一个带有“search_value”参数的函数，并希望打印“search_value”项的索引位置，包括子列表的列表索引。

示例:

grocery = ["Juice", "Tomato", "Potato", "Banana", "Milk", "Bread"]
clothes = ["Shirt", "Pant", "Jacket", "Sweater", "Hat", "Pajama", "T-Shiraz", "Polo"]

master_list = [grocery, clothes]

预期结果:

"The Item You Searched For is", search_value, ". It is in the first/second list with index position of:", index #

我是Python新手并且已经编写了工作代码。只是想知道如何以更少的努力做到这一点

def function(in_coming_string_to_search):

grocery = ["Juice", "Tomato", "Potato", "Banana", "Milk", "Bread"]
clothes = ["Shirt", "Pant", "Jacket", "Sweater", "Hat", "Pajama", "T-Shiraz", "Polo"]
master_list = [grocery, clothes]

length = int(len(master_list))
print master_list, "List has:", length, "list items", '\n'
to_do_list_first_array_index = 0
counter = 0
list_one_length = int(len(master_list[0]))

while counter < list_one_length:
    for a in master_list[to_do_list_first_array_index]:
        # print a
        if a == in_coming_string_to_search:
            print "The Item You Searched For is", in_coming_string_to_search, ". It is in the first list with index position of:", counter
        counter = counter + 1

to_do_list_second_array_index = 1
counter2 = 0
list_two_length = int(len(master_list[1]))

while counter2 < list_two_length:
    for b in master_list[to_do_list_second_array_index]:
        if b == in_coming_string_to_search:
            print "The Item You Searched For is", in_coming_string_to_search, ". It is in the second list with index position of:", counter2
        counter2 = counter2 + 1

if __name__ == '__main__':
 string_to_search = "Tomato"
 function(string_to_search)

Answer 1

怎么样(假设master_list及其子列表在全局范围内在此之前一次性定义):

def search(needle):
    for i, sublist in enumerate(master_list):
        where = sublist.find(in_coming_string_to_search)
        if where == -1: continue
        print "The Item You Searched For is", needle
        print "It is in the {} sublist, at {}".format(nd(i), where)
        return
    print "Could not find {} anywhere".format(needle)

ordinals = "first", "second", "third", "fourth", "fifth" 
def nd(i):
    if i<len(ordinals): return ordinals[i]
    return "{}th".format(i)