将词典列表转换为列表

Question

I have a list of fields in this form

fields = [{'name':'count', 'label':'Count'},{'name':'type', 'label':'Type'}]

I'd like to extract just the names and put it in a list. Currently, I'm doing this:

names = [] 
for field in fields:
    names.append(field['name'])

Is there another way to do the same thing, that doesn't involve looping through the list.

I'm using python 2.7.

Thanks for your help.!

Answer 1

You can use a list comprehension:

>>> fields = [{'name':'count', 'label':'Count'},{'name':'type', 'label':'Type'}]
>>> [f['name'] for f in fields]
['count', 'type']

Answer 2

Take a look at list comprehensions

Maybe try:

names = [x['name'] for x in fields]

Example:

>>> fields = [{'name':'count', 'label':'Count'},{'name':'type', 'label':'Type'}]
>>> [x['name'] for x in fields]
['count', 'type']

---

If names already exists you can map (creating it in the same manner) to append each member of the new list you've created:

map(names.append, [x['name'] for x in fields])

Example:

>>> a = [1, 2 ,3]
>>> map(a.append, [2 ,3, 4])
[None, None, None]
>>> a
[1, 2, 3, 2, 3, 4]

Edit:

Not sure why I didn't mention list.extend , but here is an example:

names.extend([x['name'] for x in fields])

---

or simply use the + operator:

names += [x['name'] for x in fields]

you can also filter "on the fly" using an if statement just like you'd expect:

names = [x['name'] for x in fields if x and x['name'][0] == 'a']

Example:

>>> l1 = ({'a':None}, None, {'a': 'aasd'})
>>> [x['a'] for x in l1 if (x and x['a'] and x['a'][0] == 'a')]
['aasd']

that will generate a list of all xs with names that start with 'a'

Answer 3

names = [x['name'] for x in fields] would do.

and if the list names already exists then:

names += [x['name'] for x in fields]

Answer 4

If the form is well defined, meaning, 'name' must be defined in each dict, then you may use:

map(lambda x: x['name'], fields)

else, you may use filter before extracting, just to make sure you don't get KeyError

map(lambda x: x['name'], filter(lambda x: x.has_key('name'), fields))