How to custom sort a list in python

Question

I have a list labelled Data which I want to order in the following way:

Key order = 'ID', 'GE','FN','LN','MN','EM', 'ST'

Data = ['ID:123 GE:m FN:Amir LN:Maleki EM:[email protected] MN:0400101010  ST:VIC']

I managed to order it while it is in a dictionary like this:

d= {'ID':123, 'GE':'m', 'FN':'Amir', 'LN':'Maleki', 'MN':'0400101010', 'EM':'[email protected]', 'ST':'VIC'}

keyorder = ['ID', 'GE','FN','LN','MN','EM', 'ST']

final = sorted(d.items(), key=lambda i:keyorder.index(i[0]))

print(final)

[('ID', 123), ('GE', 'm'), ('FN', 'Amir'), ('LN', 'Maleki'), ('MN', '0400101010'), ('EM', '[email protected]'), ('ST', 'VIC')]

but can I do this without making the list into a dict. If not how do i turn the list to a dict? Thank you!!

and what do you want your end-result to look like? Will it be a string? A list? A what? — Ma0
– Ma0, Commented Jan 16, 2018 at 13:33
Well you need some way to map which key corresponds with what value. — Arnav Borborah
– Arnav Borborah, Commented Jan 16, 2018 at 13:34
Is that really a list though, Data in his first message seems to be a string to be parsed. — IMCoins
– IMCoins, Commented Jan 16, 2018 at 13:47

timgeb · Accepted Answer · 2018-01-16 13:41:11Z

4

First of all, make your keyorder a dict because list.index is an O(n) operation you don't want to perform for every comparison the sorting algorithm does.

>>> keyorder = 'ID', 'GE','FN','LN','MN','EM', 'ST'
>>> keyorder = dict(zip(keyorder, range(len(keyorder))))
>>> 
>>> keyorder
{'FN': 2, 'GE': 1, 'ID': 0, 'LN': 3, 'ST': 6, 'EM': 5, 'MN': 4}

Now you can sort by splitting the part before the : from each of your strings and get the corresponding priority from keyorder.

>>> data = ['ID:123 GE:m FN:Amir LN:Maleki EM:[email protected] MN:0400101010  ST:VIC']
>>> data = data[0].split()
>>> sorted(data, key=lambda x: keyorder[x.split(':', 1)[0]])
['ID:123', 'GE:m', 'FN:Amir', 'LN:Maleki', 'MN:0400101010', 'EM:[email protected]', 'ST:VIC']

(The first two lines here sanitize your Data list because it's just a list of one element containing a string.)

answered Jan 16, 2018 at 13:41

timgeb

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2 Comments

Ma0 Over a year ago

Maybe better yet like keyorder.get(x.split(':', 1)[0], 99999999)? There might be contents in the string for which order is not important or so..

timgeb Over a year ago

@Ev.Kounis better float('inf') or float('-inf') as the fallback value if unknown prefixes can occur.

Alex · Accepted Answer · 2018-01-16 13:39:13Z

2

You can used sorted with key. It allows you to specify an arbitrary lambda expression to sort your data:

key_order = ['ID', 'GE','FN','LN','MN','EM', 'ST']
data = ['ID:12,3', 'GE:m', 'FN:Amir', 'LN:Maleki', 'EM:[email protected]', 'MN:0400101010',  'ST:VIC']
print(sorted(data, key=lambda x: key_order.index(x.split(':')[0])))

This gives as output:

['ID:12,3', 'GE:m', 'FN:Amir', 'LN:Maleki', 'MN:0400101010', 'EM:[email protected]', 'ST:VIC']

answered Jan 16, 2018 at 13:39

Alex

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1 Comment

KenjiChan1212 Over a year ago

Thank you so much I realised that I don't need to put it in a dictionary to order it. But I have to display the final result as follows ID:153 GE:m FN:John LN:Liu MN:040181010 EM:[email protected] ST:NSW

score 0 · Accepted Answer · 2018-01-16 14:55:31Z

0

Try something like this:

One line approach:

new_list=[0]*len(d)
[new_list.__setitem__(i,"{}:{}".format(j,d.get(j))) for i,j in enumerate(keyorder) if j in d]

print(new_list)

output:

['ID:123', 'GE:m', 'FN:Amir', 'LN:Maleki', 'MN:0400101010', 'EM:[email protected]', 'ST:VIC']

Detailed approach:

 d= {'ID':123, 'GE':'m', 'FN':'Amir', 'LN':'Maleki', 'MN':'0400101010', 'EM':'[email protected]', 'ST':'VIC'}

keyorder = ['ID', 'GE','FN','LN','MN','EM', 'ST']

new_list=[0]*len(d)

for i,j in enumerate(keyorder):
    if j in d:
        new_list[i]="{}:{}".format(j,d.get(j))

print(new_list)

output:

['ID:123', 'GE:m', 'FN:Amir', 'LN:Maleki', 'MN:0400101010', 'EM:[email protected]', 'ST:VIC']

edited Jan 16, 2018 at 14:55

answered Jan 16, 2018 at 14:20

user9158931

2 Comments

KenjiChan1212 Over a year ago

Hi this is great but i need to display it like this ---> ID:153 GE:m FN:John LN:Liu MN:040181010 EM:[email protected] ST:NSW

user9158931 Over a year ago

@SakiVandanaDisanayake Your name is long string ;) just kidding , Check my updated solution. And if you like my solution then don't mark it accepted , Just use it ;)

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