I have a list of lists, where the first list is a 'header' list and the rest are data lists. Each of these 'sub' lists are the same size, as shown below:
list1 = [
["Sno", "Name", "Age", "Spirit", "Status"],
[1, "Rome", 43, "Gemini", None],
[2, "Legolas", 92, "Libra", None]
]
I want to merge all of these 'sub' lists into one dictionary, using that first 'sub' list as a header row that maps each value in the row to a corresponding value in subsequent rows.
This is how my output should look:
result_dict = {
1: {"Name": "Rome", "Age": 43, "Spirit": "Gemini", "Status": None},
2: {"Name": "Legolas", "Age": 92, "Spirit": "Libra", "Status": None}
}
As you can see, 1
, 2
, etc. are unique row numbers (they correspond to the Sno
column in the header list).
So far, I am able to get every second element as the key using this code:
list1_as_dict= {p[0]:p[1:] for p in list1}
print(list1_as_dict)
Which outputs:
{
'Sno': ['Name', 'Age', 'Spirit', 'Status'],
1: ['Rome', 43, 'Gemini', None],
2: ['Legolas', 92, 'Libra', None]
}
But I don't know how make each of the data 'sub' lists a dictionary mapped to the corresponding headers.
How can I get my desired output?
CodePudding user response:
Something like
spam = [["Sno","Name","Age","Spirit","Status"],[1, "Rome", 43,"Gemini",None],[2,"Legolas", 92, "Libra",None]]
keys = spam[0][1:]
result = {sno:dict(zip(keys, values)) for sno, *values in spam[1:]}
print(result)
output
{1: {'Name': 'Rome', 'Age': 43, 'Spirit': 'Gemini', 'Status': None}, 2: {'Name': 'Legolas', 'Age': 92, 'Spirit': 'Libra', 'Status': None}}
CodePudding user response:
I'm sure you'd love Pandas. There is a bit of learning curve, but it offers a lot in exchange - most of operations over this dataset can be done in a single line of code.
import pandas as pd
pd.DataFrame(list1[1:], columns=list1[0]).set_index('Sno')
Output:
Name Age Spirit Status
Sno
1 Rome 43 Gemini None
2 Legolas 92 Libra None
CodePudding user response:
my_dict = {p[0]: dict((x,y) for x, y in zip(list1[0][1:], p[1:])) for p in list1[1:]}
print(my_dict)
Results in:
{1: {'Name': 'Rome', 'Age': 43, 'Spirit': 'Gemini', 'Status': None}, 2: {'Name': 'Legolas', 'Age': 92, 'Spirit': 'Libra', 'Status': None}}