返回与python中对应的值列表匹配的字典列表

Question

例如，这是我的字典列表：

[{'name': 'John', 'color': 'red'  },
 {'name': 'Bob',  'color': 'green'},
 {'name': 'Tom',  'color': 'blue' }] 

基于列表['blue', 'red', 'green'] 我想返回以下内容：

[{'name': 'Tom',  'color': 'blue' },
 {'name': 'John', 'color': 'red'  },
 {'name': 'Bob',  'color': 'green'}]

Answer 1

这可能有点天真，但它确实有效：

data = [
    {'name':'John', 'color':'red'},
    {'name':'Bob', 'color':'green'},
    {'name':'Tom', 'color':'blue'}
]
colors = ['blue', 'red', 'green']
result = []

for c in colors:
    result.extend([d for d in data if d['color'] == c])

print result

Answer 2

更新：

>>> list_ = [{'c': 3}, {'c': 2}, {'c': 5}]
>>> mp = [3, 5, 2]
>>> sorted(list_, cmp=lambda x, y: cmp(mp.index(x.get('c')), mp.index(y.get('c'))))
[{'c': 3}, {'c': 5}, {'c': 2}]

Answer 3

您可以使用任何自定义键功能sort。

>>> people = [
    {'name': 'John', 'color': 'red'},
    {'name': 'Bob', 'color': 'green'},
    {'name': 'Tom', 'color': 'blue'},
]
>>> colors = ['blue', 'red', 'green']
>>> sorted(people, key=lambda person: colors.index(person['color']))
[{'color': 'blue', 'name': 'Tom'}, {'color': 'red', 'name': 'John'}, {'color': 'green', 'name': 'Bob'}]

list.index 需要线性时间，因此如果颜色数量可以增加，则转换为更快的键查找。

>>> colorkeys = dict((color, index) for index, color in enumerate(colors))
>>> sorted(people, key=lambda person: colorkeys[person['color']])
[{'color': 'blue', 'name': 'Tom'}, {'color': 'red', 'name': 'John'}, {'color': 'green', 'name': 'Bob'}]

Answer 4

重复哈托的解决方案：

>>> from pprint import pprint
>>> [{'color': 'red', 'name': 'John'},
...  {'color': 'green', 'name': 'Bob'},
...  {'color': 'blue', 'name': 'Tom'}]
[{'color': 'red', 'name': 'John'}, {'color': 'green', 'name': 'Bob'}, {'color':
'blue', 'name': 'Tom'}]
>>> data = [
...     {'name':'John', 'color':'red'},
...     {'name':'Bob', 'color':'green'},
...     {'name':'Tom', 'color':'blue'}
... ]
>>> colors = ['blue', 'red', 'green']
>>> result = [d for d in data for c in colors if d['color'] == c]
>>> pprint(result)
[{'color': 'red', 'name': 'John'},
 {'color': 'green', 'name': 'Bob'},
 {'color': 'blue', 'name': 'Tom'}]
>>>

主要区别在于使用列表推导式构建结果。

编辑：我在想什么？这显然需要使用 any() 表达式：

>>> from pprint import pprint
>>> data = [{'name':'John', 'color':'red'}, {'name':'Bob', 'color':'green'}, {'name':'Tom', 'color':'blue'}]
>>> colors = ['blue', 'red', 'green']
>>> result = [d for d in data if any(d['color'] == c for c in colors)]
>>> pprint(result)
[{'color': 'red', 'name': 'John'},
 {'color': 'green', 'name': 'Bob'},
 {'color': 'blue', 'name': 'Tom'}]
>>>

Answer 5

这是一个简单的循环函数：

# Heres the people:
people = [{'name':'John', 'color':'red'}, 
          {'name':'Bob', 'color':'green'}, 
          {'name':'Tom', 'color':'blue'}] 

# Now we can make a method to get people out in order by color:
def orderpeople(order):
    for color in order:
        for person in people:
            if person['color'] == color:
                yield person

order = ['blue', 'red', 'green']
print(list(orderpeople(order)))

现在，如果您有很多人，那将非常慢。然后你可以只循环一次，但是按颜色建立一个索引：

# Here's the people:
people = [{'name':'John', 'color':'red'}, 
          {'name':'Bob', 'color':'green'}, 
          {'name':'Tom', 'color':'blue'}] 

# Now make an index:
colorindex = {}
for each in people:
    color = each['color']
    if color not in colorindex:
        # Note that we want a list here, if several people have the same color.
        colorindex[color] = [] 
    colorindex[color].append(each)

# Now we can make a method to get people out in order by color:
def orderpeople(order):
    for color in order:
        for each in colorindex[color]:
            yield each

order = ['blue', 'red', 'green']
print(list(orderpeople(order)))

即使对于非常大的列表，这也会非常快。

Answer 6

给定：

people = [{'name':'John', 'color':'red'}, {'name':'Bob', 'color':'green'}, {'name':'Tom', 'color':'blue'}]
colors = ['blue', 'red', 'green']

你可以这样做：

def people_by_color(people, colors):
  index = {}
  for person in people:
    if person.has_key('color'):
      index[person['color']] = person       
  return [index.get(color) for color in colors]

如果您要使用相同的字典列表但不同的颜色列表多次执行此操作，您需要拆分索引构建并保留索引，这样您就不需要每次都重新构建它.