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| From | jmp <jeanmichel@sequans.com> |
| Newsgroups | comp.lang.python |
| Subject | Re: Summary grid |
| Date | Wed, 22 Jun 2016 18:20:37 +0200 |
| Lines | 56 |
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On 06/22/2016 04:46 PM, Jignesh Sutar wrote:
> Say I have list of data as given in the example code below, I want to find
> all the unique categories (alphabetic letters) and unique IDs (numbers) and
> then produce a summary grid as manually entered in the "results". How could
> I code this?
>
> Many thanks in advance,
> Jignesh
>
>
> data= ["A.1", "A.2", "A.3", "B.1", "C.2", "C.3", "D.4", "E.5", "E.6"]
>
> cols=[]
> rows=[]
> for item in data:
> i=item.split(".")
> if i[0] not in cols: cols.append(i[0])
> if i[1] not in rows: rows.append(i[1])
>
> print cols
> print rows
>
> results=
> [["Row/Col", "A", "B", "C", "D", "E"],
> [1, 1, 1, 0, 0, 0],
> [2, 1, 0, 1, 0, 0],
> [3, 1, 0, 1, 0, 0],
> [4, 0, 0, 0, 1, 0],
> [5, 0, 0, 0, 0, 1],
> [6, 0, 0, 0, 0, 1]]
>
Easily done using only builtins and list comprehension:
data= ["A.1", "A.2", "A.3", "B.1", "C.2", "C.3", "D.4", "E.5", "E.6"]
# python magic with the "zip" function
# "set" will take care of removing duplicates
categories, ids = map(set, zip(*[d.split('.') for d in data]))
results = []
for id_ in sorted(map(int,ids)):
results.append([data.count("%s.%d" % (cat, id_)) for cat in
sorted(categories)])
print results
If you don't really understand the zip function:
http://nedbatchelder.com/text/iter.html
The whole presentation is worth reading.
Cheers,
Jm
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Re: Summary grid jmp <jeanmichel@sequans.com> - 2016-06-22 18:20 +0200
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