Question

我有两个包含列表的列表。我希望missing_cards中每个列表中的前两项用作master_memory中的搜索条件。如果它在master_memory中找到它们，则应打印在master_memory中找到的列表。如果在master_memory中找不到它，则会打印“不在主列表中：”，然后在missing_cards中打印该列表的前2项。

我想要打印：

不可用，位置为：a b已标记

不可用，位置为：a b收藏夹

不在主列表中：c d

不在主列表中：e f

我试过了：

card_name = 0
card_set = 1
card_location = 2
missing_cards=[['a','b','1'],['c','d','1'],['e','f','1']]
master_memory=[['a','b','tagged'],['c','e','slot_1'],['a','b','favorites'],['f','p','slot_2']


for missing_cards_row in missing_cards:
    if (missing_cards_row[card_name],missing_cards_row[card_set]) not in master_memory:
        print "Not in the Master list:",missing_cards_row[card_name],missing_cards_row[card_set]
    else:
        for master_memory_row in master_memory:
            if missing_cards_row[card_name]==master_memory_row[card_name] and missing_cards_row[card_set]==master_memory_row[card_set]:
                print "Not available, location is:",master_memory_row[card_name], master_memory_row[card_set], master_memory_row[card_location]

但这不起作用。它表示missing_cards中的所有项目都不在master_memory中，因为它在'master_memory'中找不到该列表，即使我希望它只是查看每个列表中的前两项。想法？

Answer 1

以下内容如何：

card_name = 0
card_set = 1
card_location = 2
missing_cards=[['a','b','1'],['c','d','1'],['e','f','1']]
master_memory=[['a','b','tagged'],['c','e','slot_1'],['a','b','favorites'],['f','p','slot_2']]

# Assuming missing_cards always has triplets, loop and extract each value
for x,y,z in missing_cards:

    # Remember if we had at least one match in master_memory
    found = False

    # Loop master_memory and check for existance of x y in it
    for mm in master_memory:

        # Check the first 2 elements
        if x in mm and y in mm:
            # Print matching location
            print "Not available, location is: %s" % ' '.join(mm)
            # Flag as found in master_memory
            found = True

    # Once here, check if that we had at least one match - if not print relevant message
    if not found:
        print "Not in the Master list: %s %s" % (x, y)

输出：

Not available, location is: a b tagged
Not available, location is: a b favorites
Not in the Master list: c d
Not in the Master list: e f

以上几点假设：

missing_cards总是三胞胎：

我可以将此更改为for mc in missing_cards，然后将其用作mc[0]和mc[1]进行检查，而不是x和y。这样missing_cards中的项目数可能会有所不同
匹配顺序无关紧要，即（a，b）将匹配（b，a）：

如果需要更改，则应使用x in mm代替x == mm[0] and y == mm[1]。这样我们还可以确保missing_cards项目在master_memory

Answer 2

你必须使用嵌套的for循环。

card_name = 0
card_set = 1
card_location = 2
for missing_cards_row in missing_cards:
    #This variable will keep track, if the card is found or not
    found = false; 
    for master_memory_row in master_memory:
        if(missing_cards_row[card_name]==master_memory_row[card_name] and missing_cards_row[card_set]==master_memory_row[card_set]):
             print "Not available, location is:",master_memory_row[card_name], master_memory_row[card_set], master_memory_row[card_location]
             found = true
    #This flag will inform if the missing card is found in master card or not
    if (found==false): 
        print "Not in the Master list:",missing_cards_row[card_name],missing_cards_row[card_set]

在第二个列表列表中的列表列表中查找项目

2 个答案: