将字典列表转换为新字典

Question

I'm trying to parse some html table content and I have a payload that looks like this:

"payload": [
    {
      "Cell #1": "Origin",
      "Cell #2": "Destination",
      "Cell #3": "Miles",
      "Cell #4": "Rate"
    },
    {
      "Cell #1": "Pampa, TX",
      "Cell #2": "Sallisaw, OK",
      "Cell #3": "207",
      "Cell #4": "$725"
    },
    {
      "Cell #1": "Pampa, TX",
      "Cell #2": "Sallisaw, OK",
      "Cell #3": "207",
      "Cell #4": "$725"
    },
    {
      "Cell #1": "Pampa, TX",
      "Cell #2": "Atoka, OK",
      "Cell #3": "176",
      "Cell #4": "$625"
    },
    {
      "Cell #1": "Pampa, TX",
      "Cell #2": "Wichita, KS",
      "Cell #3": "460",
      "Cell #4": "$1,150"
    }
  ]

You can see that the first dictionary in this list is the table headers. Then every item after is ordered by its corresponding header.

I am trying to make it look like this:

        [
            {
                "Origin": "Pampa, TX",
                "Destination": "Sallisaw, OK",
                "Miles": "207",
                "Rate": "$725"
            },
        ]

The catch is that the first list item which is the headers could change to add/remove "columns" so it would need to be versatile.

What I'm Trying First I believe that each dict should be converted to a list of their values:

rows = []

for row in rows:
   rows.append(row.values)

This outputs list of lists:

rows = [['Origin', 'Destination', 'Miles', 'Rate']), ['Pampa, TX', 'Sallisaw, OK', '207', '$725'], ['Pampa, TX', 'Sallisaw, OK', '207', '$725'], ['Pampa, TX', 'Atoka, OK', '176', '$625'], ['Pampa, TX', 'Wichita, KS', '460', '$1,150']]

normally if I was converting this and I knew the headers prior to receiving the payload, I would then be able to build the dictionary based on the preset headers.

Something like:

        for r in rows:
            converted_row = {
                "Origin": r[0],
                "Destination": r[1],
                "Miles": r[2],
                "Rate": r[3]
            }

But that doesn't solve for the part that the columns are dynamic. Also assuming that the ordering always matches. I have a feeling there is a more dynamic way to do this with list comprehension.

Answer 1

If dicts is your list of dictionaries and you are using at least Python 3.7 such that we can assume a deterministic order of dict items, then you can issue:

>>> keys = dicts[0].values()
>>> [dict(zip(keys, d.values())) for d in dicts[1:]]
[{'Origin': 'Pampa, TX', 'Destination': 'Sallisaw, OK', 'Miles': '207', 'Rate': '$725'}, 
 {'Origin': 'Pampa, TX', 'Destination': 'Sallisaw, OK', 'Miles': '207', 'Rate': '$725'},
 {'Origin': 'Pampa, TX', 'Destination': 'Atoka, OK', 'Miles': '176', 'Rate': '$625'},
 {'Origin': 'Pampa, TX', 'Destination': 'Wichita, KS', 'Miles': '460', 'Rate': '$1,150'}]

Answer 2

assuming you have this list stored in a variable named payload ...

you can just grab the headers from the first entry and use them as keys for the values in the remaining entries, and turn them into a list of dictionaries using a list comprehension

based on the OP I'm assuming you want a list of similar dictionaries for all entries:

headers = payload[0].values()

output = [dict(zip(headers,entry.values())) for entry in payload[1:]]

output:

[{'Destination': 'Sallisaw, OK',
  'Miles': '207',
  'Origin': 'Pampa, TX',
  'Rate': '$725'},
 {'Destination': 'Sallisaw, OK',
  'Miles': '207',
  'Origin': 'Pampa, TX',
  'Rate': '$725'},
 {'Destination': 'Atoka, OK',
  'Miles': '176',
  'Origin': 'Pampa, TX',
  'Rate': '$625'},
 {'Destination': 'Wichita, KS',
  'Miles': '460',
  'Origin': 'Pampa, TX',
  'Rate': '$1,150'}]

Answer 3

you can use this with very easy to understand code

#Declare an empty list to record the desired keys
keys=[]

#Filling the keys (dynamic and can be used for any other object)
for i in p[0].values():
    keys.append(i)

#Declare an empty ans list 
ans=[]   

#Loop over the given payload values from 2nd value to last
for i in p[1:]:

    #making the current ith dict values as a list
    ls=list(i.values())

    #declare an empty dictionary to catch the current iteration values
    dict={}

    #filling the current iteration dict values to the keys[]
    for j in range(len(ls)):
        dict[keys[j]]=ls[j]

    #finally appending the current iteration temp dict to final ans list
    ans.append(dict)

return ans