在列表中设置字典字段的pythonic方法是什么

Question

I have two lists, one containing int values:

values = [5,10,15,20]

the other is a list of dicts:

dicts = [{'a':0, 'b':0, 'c':0},
         {'a':0, 'b':0, 'c':0},
         {'a':0, 'b':0, 'c':0},
         {'a':0, 'b': 0, 'c':0}]

I want to get the first value from values list and set the field "a" of the first dict with the value coming from the "values" list and so on. In other words output should be:

dicts = [{'a':5, 'b':0, 'c':0},
         {'a':10, 'b':0, 'c':0},
         {'a':15, 'b':0, 'c':0},
         {'a':20, 'b': 0, 'c':0}]

What is the Pythonic way to do this?

Answer 1

Well you could use a zip to set the value for 'a' for every dictionary. Something like:

for dic,val in zip(dicts,values):
    dic['a'] = val

Or putting it in a one-liner :

for dic,val in zip(dicts,values): dic['a'] = val

Note that there is no difference with the previous code fragment. It depends on what your eye likes the most.

We here update the dictionaries: we do no create copies, etc. That is an important distinction since variables that could reference to the given dictionaries will reflect the changes we make.

zip(..) takes as input a sequence or more iterables (here dicts and values ) and generates tuples of elements of these iterables. So it emits:

(values[0],dicts[0]), (values[1],dicts[1]), ...

Now we do tuple-unpacking : we unify val and dic with one of the values and one of the dicts . For each of the tuples we thus set the dic['a'] value.

Console test:

$ python3
Python 3.5.2 (default, Nov 17 2016, 17:05:23) 
[GCC 5.4.0 20160609] on linux
Type "help", "copyright", "credits" or "license" for more information.
>>> values = [5,10,15,20]
>>> dicts = [{'a':0, 'b':0, 'c':0},
...          {'a':0, 'b':0, 'c':0},
...          {'a':0, 'b':0, 'c':0},
...          {'a':0, 'b': 0, 'c':0}]
>>> for dic,val in zip(dicts,values): dic['a'] = val
... 
>>> dicts
[{'b': 0, 'c': 0, 'a': 5}, {'b': 0, 'c': 0, 'a': 10}, {'b': 0, 'c': 0, 'a': 15}, {'b': 0, 'c': 0, 'a': 20}]

Answer 2

For the sake of a Pythonic one-liner (although I strongly suggest you prefer Willem's answer for clarity/readability ):

dicts = [{key: val if key != 'a' else z_val 
            for key, val in d.items()} for d, z_val in zip(dicts, values)]

dicts
Out[5]: 
[{'a': 5, 'b': 0, 'c': 0},
 {'a': 10, 'b': 0, 'c': 0},
 {'a': 15, 'b': 0, 'c': 0},
 {'a': 20, 'b': 0, 'c': 0}]

Edit: Removed the redundant if else for the dictionary keys.

Answer 3

Try this code :

values = [5,10,15,20]
dicts = [{'a':0, 'b':0, 'c':0},
         {'a':0, 'b':0, 'c':0},
         {'a':0, 'b':0, 'c':0},
         {'a':0, 'b': 0, 'c':0}]

res1 =  dict(zip(dicts[0],zip(*[d.values() for d in dicts])))# tronspose a list of dicts /// res1 = {'a': (0, 0, 0, 0), 'c': (0, 0, 0, 0), 'b': (0, 0, 0, 0)}
res1['a'] = values
res = [dict(zip(res1,t)) for t in zip(*res1.values())]# tronspose res1