将字典中的键同步到列表中的多个值

Question

i have a dict of list in following format,

mydict = 

{ a: [1, 2],
  b: [2, 2],
  c: [1, 0],
  d: [1, 1]
}

and another list with just two elements,

mylist =
    [x, y]

I read the values from mydict like this

for key, values in mydict.items():
    print key, values

this prints,

  a [1, 2]
  b [2, 2]
  c [1, 0]
  d [1, 1]

but i want the following output,

x : [1, 2, 1, 1]
y : [2, 2, 0, 1]

which means, the first column of mydict and first value of mylist makes a dictionary.

so i modified loop like this,

for key, values in mydict.items():
    newdict = {mylist[0]: mydict[0],
               mylist[1]: mydict[1] 
              }

which gives me the desired output,

x : [1, 2, 1, 1]
y : [2, 2, 0, 1]

but the problem is if the length of mylist and mydict increases then it won't work, and I need to manually add a new line,

mylist[2]: mydict[2] 

How can I make sure that keys are assigned automatically without add a new line, everytime new value in mylist is added?

PS: lenth of list value in mydict will always be equal to length of mylist .

Edit:

my output should also preserve value of keys from mydict ,

x : [1, 2, 1, 1]
y : [2, 2, 0, 1]
keys: a, b, c, d

thank you

EDIT:

        0        1
0       a        [1,2],
1       b        [2,2],
2       c        [1,0],   
3       d        [1,1] 

Answer 1

You can use zip ²

dict(zip(['x', 'y'], zip(*dct.values())))

# Result
{'x': (1, 2, 1, 1), 'y': (2, 2, 0, 1)}

This uses the zip transpose:

>>> list(zip(*dct.values()))
[(1, 2, 1, 1), (2, 2, 0, 1)]

If you want to keep your keys:

dict(zip(['key', 'x', 'y'], zip(*[(i, *j) for i, j in dct.items()])))

# Result
{'key': ('a', 'b', 'c', 'd'), 'x': (1, 2, 1, 1), 'y': (2, 2, 0, 1)}

Answer 2

If you are working with big data, you can use pandas

import pandas as pd

mydict = { 'a': [1, 2],
  'b': [2, 2],
  'c': [1, 0],
  'd': [1, 1]}

df = pd.DataFrame(mydict)

    0   1
a   1   2
b   2   2
c   1   0
d   1   1

Then you can get, eg

>>> df.columns = ['x', 'y'] # where ['x', 'y'] is your obj my_list
>>> df.to_dict('list')

{'x': [1, 2, 1, 1], 'y': [2, 2, 0, 1]}

---

To keep the keys

>>> d = df.to_dict('list')
>>> d.update({'keys':df.index.tolist()})