Python sort list of dictionaries

Question

I have a list of dictionaries:

AccountValues = [
{'portfolio_ref': 1, 'tag': 'FullInit', 'value': '20642.95', 'currency': 'USD', 'percent': 0.0}, 
{'portfolio_ref': 1, 'tag': 'FullMaint', 'value': '21350.54', 'currency': 'USD', 'percent': 0.0}, 
{'portfolio_ref': 1, 'tag': 'NetLiq', 'value': '70976.05', 'currency': 'USD', 'percent': 100.0} ]

Simple mission per SQL description: Order by portfolio_ref ASC, percent DESC

What I tried unsuccessfully:

sorted(AccountsValues, key=lambda x: (x[1],-x[4]))

which gives me

KeyError: 1

Second attempt:

import operator
result = sorted(myAccountsValues, key=itemgetter('percent'))

which fails to sort on percentage.

What about what you tried didn't work? It's always helpful to be clear about what you tried, what you expected, and what you got instead. For instance did you get an error message? Or just the wrong result? Did you make any attempts at debugging? — dimo414
– dimo414, Commented Sep 25, 2018 at 23:01
The reason your second attempt didn't seem to work is that your data is already sorted on percentage... — kindall
– kindall, Commented Sep 25, 2018 at 23:12

jpp · Accepted Answer · 2018-09-25 22:59:28Z

3

You can use dict.__getitem__ or its syntactic sugar []:

res = sorted(AccountValues, key=lambda x: (x['portfolio_ref'], -x['percent']))

Remember that dictionaries are not indexable by integers. Historically (pre-3.6), they are not even ordered. Even in Python 3.7, you cannot directly extract the nth key or value.

Result:

print(res)

[{'portfolio_ref': 1, 'tag': 'NetLiq', 'value': '70976.05', 'currency': 'USD', 'percent': 100.0},
 {'portfolio_ref': 1, 'tag': 'FullInit', 'value': '20642.95', 'currency': 'USD', 'percent': 0.0},
 {'portfolio_ref': 1, 'tag': 'FullMaint', 'value': '21350.54', 'currency': 'USD', 'percent': 0.0}]

answered Sep 25, 2018 at 22:59

jpp

166k37 gold badges301 silver badges363 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Prune · Accepted Answer · 2018-09-25 23:06:16Z

1

You just have to combine all the things you did correctly: sort keys as a tuple and the proper way of referencing a dict entry:

>>> sorted(AccountValues, key=lambda x: (x["portfolio_ref"], -x["percent"]))
[{'tag': 'NetLiq', 'portfolio_ref': 1, 'value': '70976.05', 'percent': 100.0, 'currency': 'USD'},
 {'tag': 'FullInit', 'portfolio_ref': 1, 'value': '20642.95', 'percent': 0.0, 'currency': 'USD'},
 {'tag': 'FullMaint', 'portfolio_ref': 1, 'value': '21350.54', 'percent': 0.0, 'currency': 'USD'}]

Better yet, use

sorted(AccountValues, key=itemgetter("portfolio_ref", "percent"))

Your first attempt failed because x[1] and x[4] are not valid references into the dictionaries: you have to use the labels you originally gave, not relative positions.

Your second attempt is deficient only because you don't have the secondary sort key.

edited Sep 25, 2018 at 23:06

answered Sep 25, 2018 at 23:01

Prune

78k14 gold badges63 silver badges83 bronze badges

1 Comment

vash_the_stampede Over a year ago

As I was pasting, also note @jpp posted same response, gj to both

Collectives™ on Stack Overflow

Python sort list of dictionaries

2 Answers 2

Comments

1 Comment

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

1 Comment

Your Answer

Sign up or log in

Post as a guest

Related