python - Expand a dict containing list items into a list of dict pairs

Question

If I have a dictionary containing lists in one or more of its values:

data = {
  'a':0,
  'b':1,
  'c':[0, 1, 2],
  'pair':['one','two']
}

How can I get a list of dict tuples paired by pair and iterating over c, with all else remaining constant? E.g.

output = [
    ({
        'a':0,
        'b':1,
        'c':0,
        'pair':'one'
    },
    {
        'a':0,
        'b':1,
        'c':0,
        'pair':'two'
    }),
    ({
        'a':0,
        'b':1,
        'c':1,
        'pair':'one'
    },
    ...
]

标签： pythonpython-2.7dictionarylist-comprehensiondictionary-comprehension

Answer 1

好吧，这感觉不是特别优雅，但您可以使用嵌套的 for 循环或列表推导：

output = []
for i in data['c']:
  output.append(tuple({'a': 0, 'b': 1, 'c': i, 'pair': p} for p in data))

或者

output = [tuple({'a': 0, 'b': 1, 'c': i, 'pair': p} for p in data['pair']) for i in data['c']]

更简洁的解决方案可能会将组件 dict 的生成分离到一个函数中，如下所示：

def gen_output_dict(c, pair):
  return {'a': 0, 'b': 1, 'c': c, 'pair': pair}

output = []
for i in data['c']:
  output.append(tuple(gen_output_dict(i, p) for p in data['pair']))