Sorting a list

[Peter Otten]

DFS wrote:

> cntText = 60
> cntBool = 20
> cntNbrs = 30
> cntDate = 20
> cntBins = 20
> 
> strText = "  text:     "
> strBool = "  boolean:  "
> strNbrs = "  numeric:  "
> strDate = "  date-time:"
> strBins = "  binary:   "
> 
> colCounts = [(cntText,strText) , (cntBool,strBool), (cntNbrs,strNbrs) ,
> (cntDate,strDate) , (cntBins,strBins)]
> 
> # sort by alpha, then by column type count descending
> colCounts.sort(key=lambda x: x[1])
> colCounts.sort(key=lambda x: x[0], reverse=True)
> for key in colCounts: print key[1], key[0]]
> 
> -------------------------------------------------
> 
> Output (which is exactly what I want):
> 
>    text:      60
>    numeric:   30
>    binary:    20
>    boolean:   20
>    date-time: 20
> 
> -------------------------------------------------
> 
> 
> But, is there a 1-line way to sort and print?

Yes, but I would not recommend it. You can replace the sort() method 
invocations with nested calls of sorted() and instead of

for item in items:
    print convert_to_str(item)

use

print "\n".join(convert_to_str(item) for item in items)

Putting it together:

>>> from operator import itemgetter as get
>>> print "\n".join("{1} {0}".format(*p) for p in sorted(
... sorted(colCounts, key=get(1)), key=get(0), reverse=True))
  text:      60
  numeric:   30
  binary:    20
  boolean:   20
  date-time: 20

You could also cheat and use

lambda v: (-v[0], v[1])

and a single sorted().