Question

我有一个带有类别的单词词典。如果单词与列表中的单词匹配，我想输出类别。这就是我的代码目前的样子：

dictionary = {
    "object" : 'hat',
    "animal" : 'cat',
    "human" : 'steve'
}

list_of_things = ["steve", "tom", "cat"]

for categories,things in dictionary.iteritems():
    for stuff in list_of_things:

        if stuff in things:
            print categories
        if stuff not in things:
            print "dump as usual"

目前输出如下：

dump as usual
dump as usual
dump as usual
human
dump as usual
dump as usual
dump as usual
dump as usual
animal

但我希望输出看起来像这样：

human
dump as usual 
animal

我不希望我的列表打印它在字典中迭代的所有内容。我只希望它打印匹配的术语。我该怎么做？

Answer 1

您可以在内部for循环中使用布尔值，该布局从False（未找到类别）更改为True（找到类别），然后仅在处打印categories for循环if boolean = False:

的结束

isMatchFound = False
for categories, things in dictionary.iteritems():
    isMatchFound = False
    for stuff in list_of_things:
        if stuff in things:
            isMatchFound = True
            print stuff
    if isMatchFound == False:
        print("dump as usual")

Answer 2

根据您的实际数据的大小，您可以执行

category_thing= {
    "object" : 'hat',
    "animal" : 'cat',
    "human" : 'steve'
}

list_of_things = ["steve", "tom", "cat"]
thingset=set(list_of_things) # just for speedup

for category,thing in category_thing.iteritems():

        if thing in thingset:
            print category
        else:
            print "dump as usual"

或者如果你的映射真的像你的例子那样简单，你可以做到

category_thing= {
    "object" : 'hat',
    "animal" : 'cat',
    "human" : 'steve'
}

thing_category=dict((t,c) for c,t in category_thing.items()) # reverse the dict - if you have duplicate values (things), you should not do this

list_of_things = ["steve", "tom", "cat"]

for stuff in list_of_things:
     msg=thing_category.get(stuff,"dump as usual")
     print msg

Answer 3

首先，你的词典结构很差;似乎交换了键和值。通过将类别用作键，每个类别只能有一个对象，这可能不是您想要的。这也意味着您必须阅读字典中的每个条目才能查找项目，这通常是一个不好的标志。对此的修复很简单：将项目放在冒号的左侧，将类别放在右侧。然后，您可以使用“in”运算符轻松搜索字典。

就你直接提出的问题而言，你应首先循环遍历list_of_things，检查每个字典，然后打印结果。这将在列表中的每个项目上打印一件事。

dictionary = {
    'hat' : 'object',
    'cat' : 'animal',
    'steve' : 'human'
}

list_of_things = ['steve', 'tom', 'cat']

for thing in list_of_things:
    if thing in dictionary:
        print dictionary[thing]
    else:
        print "dump as usual"

输出：

human
dump as usual
animal

Answer 4

由于输出中只需要3行，因此应重新排序for循环。

for stuff in list_of_things:
  print_val = None
  for categories,things in dictionary.iteritems():
    if stuff in things:
      print_val=categories
  if print_val is None:
    print_val="dump as usual"
  print print_val

将字典与列表匹配（Python）

4 个答案: