python sort data structure by value lowest to highest

Question

I'd like to sort a list of tuples by the magnitude of a unicode string, i.e. the first member of the tuple.

The "raw" tuples look like this:

A

[

   [
        (u'90000', 100318), 
        (u'21000', 58094), 
        (u'50000', 14695), 
        (u'250000', 8190), 
        (u'100000', 5718), 
        (u'40000', 4276)
     ]
]

After processing I'd like it to look like this:

B

[

   [
        (u'250000', 8190), 
        (u'100000', 5718), 
        (u'90000', 100318),
        (u'50000', 14695), 
        (u'40000', 4276),
        (u'21000', 58094)
     ]
]

I've tried some different approaches but I've not been able to get it to persist.

How to sort the list of tuples in A to render it like B?

Did you just copy-paste an email or chat conversation or something into the question box? — user2357112
– user2357112, Commented Sep 17, 2017 at 0:24
I don't see how the linked question is related or more general. Here you're sorting a list of tuples, there data is being read from a list of dictionaries — Nick is tired
– Nick is tired, Commented Sep 17, 2017 at 0:26

Community · Accepted Answer · 2020-06-20 09:12:55Z

1

Here is an interactive demonstration using list.sort().

>>> help(list.sort)
Help on built-in function sort:

sort(...)
    L.sort(cmp=None, key=None, reverse=False) -- stable sort *IN PLACE*;
    cmp(x, y) -> -1, 0, 1

The built-in help is useful but for a more complete explanation of the parameters we need to consult the python documentation

cmp specifies a custom comparison function of two arguments (iterable elements) which should return a negative, zero or positive number depending on whether the first argument is considered smaller than, equal to, or larger than the second argument: cmp=lambda x,y: cmp(x.lower(), y.lower()). The default value is None.

key specifies a function of one argument that is used to extract a comparison key from each list element: key=str.lower. The default value is None (compare the elements directly).

reverse is a boolean value. If set to True, then the list elements are sorted as if each comparison were reversed.

Continuing on:

>>> l = [[(u'90000', 100318),(u'21000', 58094),(u'50000', 14695),(u'250000', 8190),(u'100000', 5718),(u'40000', 4276)]]
>>> l[0].sort(key=lambda x:int(x[0]),reverse=True)
>>> l
[[(u'250000', 8190), (u'100000', 5718), (u'90000', 100318), (u'50000', 14695), (u'40000', 4276), (u'21000', 58094)]]

edited Jun 20, 2020 at 9:12

CommunityBot

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answered Sep 17, 2017 at 0:27

jq170727

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3 Comments

Nick is tired Over a year ago

You could improve this answer by actually explaining how the code works and what a sort key is

jq170727 Over a year ago

Added doc link and excerpt

Nick is tired Over a year ago

that's much better!

Often Lin · Accepted Answer · 2017-09-19 08:26:11Z

0

Due to inner list, you need to dig into next layer => A[0]

Then use sort function with key & reverse to help you achieve the goal

list.sort(A[0], key= lambda x: int(x[0]), reverse=True)

answered Sep 19, 2017 at 8:26

Often Lin

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