Iterate through a dictionary of lists one "line" at a time

[wswilson]

Here is my code:

listing = {'id': ['a', 'b', 'c'], 'name': ['Joe', 'Jane', 'Bob']}

I need to output:

id name
a Joe
b Jane
c Bob

I could do:

print 'id', 'name'
for id, name in zip(listing['id'], listing['name']): print id, name

but that only works if there are two entries in the dictionary, id and
name, and I know what they are. My problem is I don't know how many of
these entries there will be. Thanks for any help you can give!

 

[Steve Holden]

Here is my code:

listing = {'id': ['a', 'b', 'c'], 'name': ['Joe', 'Jane', 'Bob']}

I need to output:

id name
a Joe
b Jane
c Bob

I could do:

print 'id', 'name'
for id, name in zip(listing['id'], listing['name']): print id, name

but that only works if there are two entries in the dictionary, id and
name, and I know what they are. My problem is I don't know how many of
these entries there will be. Thanks for any help you can give!

>>> listing = {'id': ['a', 'b', 'c'], 'name': ['Joe', 'Jane', 'Bob']}
>>> for (id, name) in zip(listing['id'], listing['name']):

... print id, name
...
a Joe
b Jane
c Bob
regards
Steve

 

[Arnaud Delobelle]

Here is my code:

listing = {'id': ['a', 'b', 'c'], 'name': ['Joe', 'Jane', 'Bob']}

I need to output:

id name
a Joe
b Jane
c Bob

I could do:

print 'id', 'name'
for id, name in zip(listing['id'], listing['name']): print id, name

but that only works if there are two entries in the dictionary, id and
name, and I know what they are. My problem is I don't know how many of
these entries there will be. Thanks for any help you can give!

You can use zip(*sequence_of_sequences)
eg zip(*[[1,2,3],[4,5,6]) returns [[1,4], [2,5], [3,6]]
(You can think of it as a transposition function)

For example:

keys, item_lists = zip(*listing.iteritems())
print " ".join(keys)
for items in zip(*item_lists):
print " ".join(items)

would work.

HTH

 

[wswilson]

Here is my code:

listing = {'id': ['a', 'b', 'c'], 'name': ['Joe', 'Jane', 'Bob']}

I need to output:

id name
a Joe
b Jane
c Bob

I could do:

print 'id', 'name'
for id, name in zip(listing['id'], listing['name']): print id, name

but that only works if there are two entries in the dictionary, id and
name, and I know what they are. My problem is I don't know how many of
these entries there will be. Thanks for any help you can give!

You can use zip(*sequence_of_sequences)
eg zip(*[[1,2,3],[4,5,6]) returns [[1,4], [2,5], [3,6]]
(You can think of it as a transposition function)

For example:

keys, item_lists = zip(*listing.iteritems())
print " ".join(keys)
for items in zip(*item_lists):
print " ".join(items)

would work.

HTH

That works perfectly. Thanks so much.

 

[Maric Michaud]

wswilson a écrit :

Here is my code:

listing = {'id': ['a', 'b', 'c'], 'name': ['Joe', 'Jane', 'Bob']}

I need to output:

id name
a Joe
b Jane
c Bob

I could do:

print 'id', 'name'
for id, name in zip(listing['id'], listing['name']): print id, name

but that only works if there are two entries in the dictionary, id and
name, and I know what they are. My problem is I don't know how many of
these entries there will be. Thanks for any help you can give!

The most simple and generic I see, it could be even more simple if you
don't care of memory usage :

Let's fill a random dict :

In [118]: d=dict(zip((str(e) for e in xrange(10)), ([i**e for e in
xrange(5)] for i in xrange(10))))

In [119]: d
Out[119]:
{'0': [1, 0, 0, 0, 0],
'1': [1, 1, 1, 1, 1],
'2': [1, 2, 4, 8, 16],
'3': [1, 3, 9, 27, 81],
'4': [1, 4, 16, 64, 256],
'5': [1, 5, 25, 125, 625],
'6': [1, 6, 36, 216, 1296],
'7': [1, 7, 49, 343, 2401],
'8': [1, 8, 64, 512, 4096],
'9': [1, 9, 81, 729, 6561]}

go on :

In [146]: sorted_keys = tuple(sorted(d.keys()))

In [147]: from itertools import izip, chain

In [148]: sorted_keys = tuple(sorted(d.keys()))

In [149]: sorted_values = ( d[k] for k in sorted_keys )

In [150]: for vals in chain([sorted_keys], izip(*sorted_values)) :
.....: print '%5s'*len(d) % vals
.....:
.....:
0 1 2 3 4 5 6 7 8 9
1 1 1 1 1 1 1 1 1 1
0 1 2 3 4 5 6 7 8 9
0 1 4 9 16 25 36 49 64 81
0 1 8 27 64 125 216 343 512 729
0 1 16 81 256 625 1296 2401 4096 6561





--
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Maric Michaud
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