如何用一系列值替换 Python 字典的所有值？

Question

I have the following dictionary:

mydict = {('a', 'b'): 28.379,
          ('c', 'd'): 32.292,
          ('e', 'f'): 61.295,
          ('g', 'h'): 112.593,
          ('i', 'j'): 117.975}

And I would like to replace all the values with a range from 1 to 5, but keep the order of the keys. As a result, I would get this:

mydict = {('a', 'b'): 1,
          ('c', 'd'): 2,
          ('e', 'f'): 3,
          ('g', 'h'): 4,
          ('i', 'j'): 5}

The length of the dictionary is actually 22000, so I need a range from 1 to 22000. How can I do it?

Thanks in advance.

Answer 1

Using enumerate to iterate on the keys, you can do:

mydict = {('a', 'b'): 28.379,
          ('c', 'd'): 32.292,
          ('e', 'f'): 61.295,
          ('g', 'h'): 112.593,
          ('i', 'j'): 117.975}

for i, key in enumerate(mydict):  # iterates on the keys
    mydict[key] = i


print(mydict)
# {('a', 'b'): 0, ('c', 'd'): 1, ('e', 'f'): 2, ('g', 'h'): 3, ('i', 'j'): 4}

Important note: dicts are only officially ordered since Python 3.7 (and in the CPython implementation since 3.6), so this would n't make much sense with older versions of Python.

---

To answer your comment: enumerate takes an optional second parameter start (that defaults to 0)

So, if you want to start at 1, just do:

for i, key in enumerate(mydict, start=1):  # iterates on the keys
    mydict[key] = i

Answer 2

The most simple is to create another dictionary from the keys of the previous one.

mydict2=dict()
for i,key in enumerate(mydict):
    mydict2[key]=i+1

Answer 3

You can do this with a one-liner which is more compact:

mydict = {('a', 'b'): 28.379,
          ('c', 'd'): 32.292,
          ('e', 'f'): 61.295,
          ('g', 'h'): 112.593,
          ('i', 'j'): 117.975}

{k: i for i, (k, v) in enumerate(mydict.items())}

Answer 4

Pandas solution for this:

import pandas as pd
a = pd.DataFrame(mydict, index=[0]).T
a[0] = list(range(0,len(a)))
a.to_dict()[0]

#  {('a', 'b'): 0, ('c', 'd'): 1, ('e', 'f'): 2, ('g', 'h'): 3, ('i', 'j'): 4}

Answer 5

This can be done gracefully with dict.update and itertools.count , and explicit loops can be avoided:

>>> mydict = {('a', 'b'): 28.379,
...           ('c', 'd'): 32.292,
...           ('e', 'f'): 61.295,
...           ('g', 'h'): 112.593,
...           ('i', 'j'): 117.975}
>>> from itertools import count
>>> mydict.update(zip(mydict, count(1)))
>>> mydict
{('a', 'b'): 1, ('c', 'd'): 2, ('e', 'f'): 3, ('g', 'h'): 4, ('i', 'j'): 5}