Pythonic way to condese my array

[Dean Card]

I have a list of dictionaries where each dictionary defines, among other
things, a row and column number. So, my list might look like this:
[{'row':1, 'column':1, otherdata}, {'row':1, 'column':2, 'otherdata}...]

This data is passed to flash and used there to create a grid of objects that
are placed based on the row and column values.

For a given set of data I may have values column values in the range of 1, 9
inclusive.

What I would like to do is take my list of dictionaries (like the one listed
above) and shift the column numbers down to fill in missing values...
example (only the column key,value pair is show for simplification):

[{'column':1}, {'column':2}, {'column', 4}, {'column': 8}, {'column':2},
{'column', 4}]

If this were the dataset... I would want to change all 4s to 3 and all 8s to
4. That way I end up with 1 ... 4 instead of 1, 2, 4, and 8.

Is there some slick way to do this?

Thanks

[Robert Kern]

Dean Card wrote:

I have a list of dictionaries where each dictionary defines, among other
things, a row and column number. So, my list might look like this:
[{'row':1, 'column':1, otherdata}, {'row':1, 'column':2, 'otherdata}...]

This data is passed to flash and used there to create a grid of objects that
are placed based on the row and column values.

For a given set of data I may have values column values in the range of 1, 9
inclusive.

What I would like to do is take my list of dictionaries (like the one listed
above) and shift the column numbers down to fill in missing values...
example (only the column key,value pair is show for simplification):

[{'column':1}, {'column':2}, {'column', 4}, {'column': 8}, {'column':2},
{'column', 4}]

If this were the dataset... I would want to change all 4s to 3 and all 8s to
4. That way I end up with 1 ... 4 instead of 1, 2, 4, and 8.

Is there some slick way to do this?

Untested, but intended for Python 2.4:

list_o_dicts = [{'column': 1}, ...]
columns = set(x['column'] for x in list_o_dicts)
column_map = dict((column, i+1) for i, column in enumerate(sorted(columns)))
for d in list_o_dicts:
d['column'] = column_map[d['column']]

Of course, code that I wouldn't get fired for would have a slightly greater line
count and fewer one-letter variable names.

--
Robert Kern

"I have come to believe that the whole world is an enigma, a harmless enigma
that is made terrible by our own mad attempt to interpret it as though it had
an underlying truth."
-- Umberto Eco

[James Stroud]

Dean Card wrote:

I have a list of dictionaries where each dictionary defines, among other
things, a row and column number. So, my list might look like this:
[{'row':1, 'column':1, otherdata}, {'row':1, 'column':2, 'otherdata}...]

This data is passed to flash and used there to create a grid of objects that
are placed based on the row and column values.

For a given set of data I may have values column values in the range of 1, 9
inclusive.

What I would like to do is take my list of dictionaries (like the one listed
above) and shift the column numbers down to fill in missing values...
example (only the column key,value pair is show for simplification):

[{'column':1}, {'column':2}, {'column', 4}, {'column': 8}, {'column':2},
{'column', 4}]

If this were the dataset... I would want to change all 4s to 3 and all 8s to
4. That way I end up with 1 ... 4 instead of 1, 2, 4, and 8.

Is there some slick way to do this?

Thanks

Not sure how slick this is, but I would do it thus:

dicts = [{'column':1}, {'column':2}, {'column': 4}, {'column': 8},
{'column':4}]

vals = list(set(d['column'] for d in dicts))
vals.sort()
amap = dict((b,a) for (a,b) in enumerate(vals))

for d in dicts:
d['column'] = amap[d['column']] + 1
The "1" in the last line comes from your requirement to start numbering
at 1 instead of 0.

--
James Stroud
UCLA-DOE Institute for Genomics and Proteomics
Box 951570
Los Angeles, CA 90095

http://www.jamesstroud.com/

[Wildemar Wildenburger]

Robert Kern wrote:

Of course, code that I wouldn't get fired for would have a slightly greater line
count and fewer one-letter variable names.

Great :D!
That one made me laugh!

wildemar