simple way to un-nest (flatten?) list

[djc]

There is I am sure an easy way to do this, but I seem to be brain dead
tonight. So:

I have a table such that I can do

[line for line in table if line[7]=='JDOC']
and
[line for line in table if line[7]=='Aslib']
and
[line for line in table if line[7]=='ASLIB']
etc

I also have a dictionary
r= {'a':('ASLIB','Aslib'),'j':('JDOC', 'jdoc')}
so I can extract values
r.values()
[('ASLIB', 'Aslib'), ('JDOC', 'jdoc')]

I would like to do

[line for line in table if line[7] in ('JDOC','jdoc','Aslib','ASLIB')]

so how should I get from
{'a':('ASLIB','Aslib'),'j':('JDOC','jdoc')}
to
('Aslib','ASLIB','JDOC','jdoc')

--
djc

[George Sakkis]

djc wrote:

There is I am sure an easy way to do this, but I seem to be brain dead
tonight. So:

I have a table such that I can do

[line for line in table if line[7]=='JDOC']
and
[line for line in table if line[7]=='Aslib']
and
[line for line in table if line[7]=='ASLIB']
etc

I also have a dictionary
r= {'a':('ASLIB','Aslib'),'j':('JDOC', 'jdoc')}
so I can extract values
r.values()
[('ASLIB', 'Aslib'), ('JDOC', 'jdoc')]

I would like to do

[line for line in table if line[7] in ('JDOC','jdoc','Aslib','ASLIB')]

so how should I get from
{'a':('ASLIB','Aslib'),'j':('JDOC','jdoc')}
to
('Aslib','ASLIB','JDOC','jdoc')

Meet itertools:

from itertools import chain
names = set(chain(*r.itervalues()))
print [line for line in table if line[7] in names]
George

[Steven D'Aprano]

On Sun, 05 Nov 2006 21:43:33 +0000, djc wrote:

There is I am sure an easy way to do this, but I seem to be brain dead
tonight. So:

I have a table such that I can do

[line for line in table if line[7]=='JDOC']
and
[line for line in table if line[7]=='Aslib']
and
[line for line in table if line[7]=='ASLIB']
etc

I also have a dictionary
r= {'a':('ASLIB','Aslib'),'j':('JDOC', 'jdoc')}
so I can extract values
r.values()
[('ASLIB', 'Aslib'), ('JDOC', 'jdoc')]

I would like to do

[line for line in table if line[7] in ('JDOC','jdoc','Aslib','ASLIB')]

What is the purpose of the "if line[7]" bit?

so how should I get from
{'a':('ASLIB','Aslib'),'j':('JDOC','jdoc')}
to
('Aslib','ASLIB','JDOC','jdoc')

Assuming you don't care what order the strings are in:

r = {'a':('ASLIB','Aslib'),'j':('JDOC','jdoc')}
result = sum(r.values(), ())

If you do care about the order:

r = {'a':('ASLIB','Aslib'),'j':('JDOC','jdoc')}
keys = r.keys()
keys.sort()
result = []
for key in keys:
result.extend(r[key])
result = tuple(result)
--
Steven.

[djc]

George Sakkis wrote:

Meet itertools:

from itertools import chain
names = set(chain(*r.itervalues()))
print [line for line in table if line[7] in names]

Steven D'Aprano wrote:

Assuming you don't care what order the strings are in:

r = {'a':('ASLIB','Aslib'),'j':('JDOC','jdoc')}
result = sum(r.values(), ())

If you do care about the order:

r = {'a':('ASLIB','Aslib'),'j':('JDOC','jdoc')}
keys = r.keys()
keys.sort()
result = []
for key in keys:
result.extend(r[key])
result = tuple(result)

Thank you everybody.
As it is possible that the tuples will not always be the same word in
variant cases
result = sum(r.values(), ())
will do fine and is as simple as I suspected the answer would be.

--
djc

[bearophileHUGS]

djc:

As it is possible that the tuples will not always be the same word in
variant cases
result = sum(r.values(), ())
will do fine and is as simple as I suspected the answer would be.

It is simple, but I suggest you to take a look at the speed of that
part of your code into your program. With this you can see the
difference:

from time import clock
d = dict((i,range(300)) for i in xrange(300))

t = clock()
r1 = sum(d.values(), [])
print clock() - t

t = clock()
r2 = []
for v in d.values(): r2.extend(v)
print clock() - t

assert r1 == r2

Bye,
bearophile

[djc]

be************@lycos.com wrote:

It is simple, but I suggest you to take a look at the speed of that
part of your code into your program. With this you can see the
difference:

from time import clock
d = dict((i,range(300)) for i in xrange(300))

t = clock()
r1 = sum(d.values(), [])
print clock() - t

t = clock()
r2 = []
for v in d.values(): r2.extend(v)
print clock() - t

Yes, interesting, and well worth noting

1 for v in d.values(): r1.extend(v)

2 from itertools import chain
set(chain(*d.itervalues()))

3 set(v for t in d.values() for v in t)

4 sum(d.values(), [])

5 reduce((lambda l,v: l+v), d.values())

on IBM R60e [CoreDuo 1.6MHz/2GB]
d = dict((i,range(x)) for i in xrange(x))
x t1 t2 t3 t4 t5
300 0.0 0.02 0.04 0.31 0.32
500 0.01 0.09 0.1 1.67 1.69
1000 0.02 0.3 0.4 16.17 16.15
0.03 0.28 0.42 16.37 16.31
1500 0.03 0.76 0.94 57.05 57.13
2000 0.07 1.2 1.66 136.6 136.97
2500 0.11 2.34 2.64 268.44 268.85

but on the other hand, as the intended application is a small command
line app where x is unlikely to reach double figures and there are only
two users, myself included:
d =
{'a':['ASLIB','Aslib'],'j':['JDOC','jdoc'],'x':['test','alt','3rd'],'y':['single',]}
0.0 0.0 0.0 0.0 0.0

And sum(d.values(), []) has the advantage of raising a TypeError in the
case of a possible mangled input.

{'a':['ASLIB','Aslib'],'j':['JDOC','jdoc'],'x':['test','alt','3rd'],'y':'single'}
r1
['ASLIB', 'Aslib', 'test', 'alt', '3rd', 'JDOC', 'jdoc', 's', 'i', 'n',
'g', 'l', 'e']
r2
set(['Aslib', 'JDOC', 'g', '3rd', 'i', 'l', 'n', 'ASLIB', 's', 'test',
'jdoc', 'alt', 'e'])
r4 = sum(d.values(), [])
TypeError: can only concatenate list (not "str") to list

--
djc