how to sort list of dictionaries by value in python

Question

I have the original variable:

files = {'1.txt' : [{'Line number': 1, 'file number': 1},
                   {'Line number': 2, 'file number': 1}],
         '2.txt' : [{'Line number': 1, 'file number': 2}]
        }

I need to sort the files by 'Line number' (although reverse, from least to the most, like this:

2.txt
total lines 1
Line number 1 file number 2
1.txt
total lines 2
Line number 1 file number 1
Line number 2 file number 1

I was writing code, but I have stuck with many lines and made it very complex. Any idea about simpler few lines of code?

Please edit your question and post what you have tried so far. — MattDMo
– MattDMo, Commented Nov 5, 2022 at 18:29

PieCot · Accepted Answer · 2022-11-05 19:00:00Z

Something like this?

sorted_files = {
    f: sorted(ll, key=lambda x: x['Line number'])
    for f, ll in sorted(files.items(), key=lambda x: len(x[1]))
}

Printing sorted_files you get:

for f, ll in sorted_files.items():
    print(f'{f}\ntotal lines {len(ll)}')
    for l in ll:
        print(' '.join([f'{k} {v}' for k, v in l.items()]))

2.txt
total lines 1
Line number 1 file number 2
1.txt
total lines 3
Line number 1 file number 1
Line number 2 file number 1
Line number 3 file number 4

Note: starting from Python 3.6 onwards, the dict type keeps the order of the insertion. So, iterating through the items of sorted_files, you will always get the right sequence of items. If you are using an older version of Python, you should not use a dict if you want to keep the order of the items, but instead, you should use a tuple of tuples or a list of lists (or any combination of list and tuple). So, you should use the following code:

sorted_files = [
    (f, sorted(ll, key=lambda x: x['Line number']))
    for f, ll in sorted(files.items(), key=lambda x: len(x[1]))
]

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