Question

我正在使用pyton 3.2 我有以下字典：

a = {"fruits":["apple","mangoes","grapes"],"colour":["apple","orange","grapes"],"number":["1","2","3"]}

我需要迭代这个字典列表，我想创建一个新的字典b，其中fruits中的值与颜色的值相同：

 b = {"fruits":["apple","grapes"],"colour":["apple", "grapes"],"number":["1","3"]}

我认为这可行，但从此失去了：

b = {}
for item in a:
   if x in a[item]:

......不知道现在该怎么办？

如何在不使用itertools的情况下完成此操作？除了我列出的字典之外，是否还有一个可用于任何字典的通用功能？

Answer 1

在这种情况下，您最好使用set：

b = {"fruits": list(set(a['fruits']).intersection(a['colour'])}
b['colour'] = b['fruits']  # or maybe make a copy here...

Answer 2

这样的东西？

a = {"fruits":["apple","mangoes","grapes"],"colour":["apple","orange","grapes"]}
b = {}

b['fruits'], b['colour'] = [], []

for fruit in a['fruits']:
    if fruit in a['colour']:
        b['fruits'].append(fruit)
        b['colour'].append(fruit)

Answer 3

如果您不想在结果字典中使用数字，可以这样做......

<强>代码：

a = {"fruits":["apple","mangoes","grapes"],"colour":["apple","orange","grapes"],"number":["1","2","3"]}

b = dict()
b['fruits'] = b['colour'] = [x for x in a['fruits'] if x in a['colour']]

print b

<强>输出：

{'colour': ['apple', 'grapes'], 'fruits': ['apple', 'grapes']}

希望这会有所帮助：）

Answer 4

b = {}
for fruit in a['fruits']:
    if fruit in a['colour']:
        b.setdefault('fruits', [])   # set if key is not present
        b.setdefault('colour', [])
        b.setdefault('number', [])
        b['fruits'].append(fruit)
        b['colour'].append(fruit)
        b['number'].append(a['number'][a['fruits'].index(fruit)])

迭代字典列表

4 个答案: