How to append dictionary to a list in loop

Question

I have at my disposal huge amount of data, in the form of a list of tuples. Each tuple has a specified format like (a, b, c, d, e). The list of tuples looks like:

tupleList = [('a1', 'b1', 'c1', 'd1', 'e1'),
             ('a2', 'b2', 'c2', 'd2', 'e2'),
             ...
             ('a10000', 'b10000', 'c10000', 'd10000', 'e100000')]

What I want is, to convert each of these tuples to a dictionary, and append the dictionary to a a final list of dictionaries. Can all this be done in a loop? The final list of dictionaries should look like:

finalDictList = [{'key1': 'a1', 'key2': 'b1', 'key3': 'c1', 'key4': 'd1', 'key5': 'e1'},
                 {'key1': 'a2', 'key2': 'b2', 'key3': 'c2', 'key4': 'd2', 'key5': 'e2'},
                 {'key1': 'a3', 'key2': 'b3', 'key3': 'c3', 'key4': 'd3', 'key5': 'e3'},
                 ...
                 {'key1': 'a10000', 'key2': 'b10000', 'key3': 'c10000', 'key4': 'd10000', 'key5': 'e10000'}]

The format of the tuples is fixed. I want to compare afterwords, value of each key of a dictionary with all others. This is why the conversion of tuple to dictionary made sense to me. Please correct me if the design paradigm itself seems wrong. Also, there are >10000 tuples. Declaring that many dictionaries is just not done.

Is there anyway to append dictionary to a list in a loop? Also, if that is possible, can we access each dictionary by it's key values, say, like finalDictList[0]['key1']?

juliomalegria · Accepted Answer · 2012-01-20 20:27:54Z

10

We're going to mix three important concepts to make this code really small and beautiful. First, a list comprehension, then, the zip method, and finally, the dict method, to build a dictionary out of a list of tuples:

my_list = [('a1', 'b1', 'c1', 'd1', 'e1'), ('a2', 'b2', 'c2', 'd2', 'e2')]
keys = ('key1', 'key2', 'key3', 'key4', 'key5')
final = [dict(zip(keys, elems)) for elems in my_list]

After that, the value of the final variable is:

>>> final
[{'key3': 'c1', 'key2': 'b1', 'key1': 'a1', 'key5': 'e1', 'key4': 'd1'},
{'key3': 'c2', 'key2': 'b2', 'key1': 'a2', 'key5': 'e2', 'key4': 'd2'}]

Also, you can get elements of a certain dictionary using the position of the dictionary in the list and the key you're looking for, i.e.:

>>> final[0]['key1']
'a1'

edited Jan 20, 2012 at 20:27

answered Jan 20, 2012 at 20:19

juliomalegria

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1 Comment

juliomalegria Over a year ago

@sneha: glad I helped, if this post solved your problem, you could choose it as accepted (green check on the left)

juliomalegria · Accepted Answer · 2013-03-24 18:34:41Z

6

Use zip to combine a pre-defined list of key names with each tuple in your input list, then pass the results to dict to make them into dicts. Wrap the whole thing in a list comprehension to process them all in one batch:

keys = ('key1', 'key2', 'key3', 'key4', 'key5')
finalDictList = [dict(zip(keys, values)) for values in tupleList]

edited Mar 24, 2013 at 18:34

juliomalegria

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answered Jan 20, 2012 at 20:19

Kirk Strauser

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Comments

Greg Hewgill · Accepted Answer · 2012-01-20 20:18:58Z

4

I'm not sure I see why you need to convert everything to a dictionary, when you've already got a list of tuples.

>>> tupleList = [('a1', 'b1', 'c1', 'd1', 'e1'),
...              ('a2', 'b2', 'c2', 'd2', 'e2'),
...              ('a10000', 'b10000', 'c10000', 'd10000', 'e100000')]
>>> [x[1] for x in tupleList]
['b1', 'b2', 'b10000']

Using Python's list comprehension syntax, you can get a list of all the n-th elements of each tuple.

answered Jan 20, 2012 at 20:18

Greg Hewgill

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3 Comments

sneha Over a year ago

I know of this, but the entries in the tuple are related to each other - that is, (a,b,c,d,e) are values that make sense together only. I understand that for comparing this is fine, but i have to work on the complete tuple in case during comparison a match occurs/doesn't occur. I can make this work, but it'll be in a round about way. Thank you! :)

Greg Hewgill Over a year ago

@sneha: Accessing elements of a tuple is almost exactly the same syntax as accessing elements of a dictionary: x[2] vs x["key2"]. If you want, you can define constants like KEY2 = 2, so you can do x[KEY2]. Converting to a list of dictionaries will take up a lot more memory for little benefit.

sneha Over a year ago

Hmmm...I guess you have a pretty good point there. Let me reconsider my design. :)

g.d.d.c · Accepted Answer · 2012-01-20 20:20:09Z

3

If the fields are fix you can do this:

fields = ['key1', 'key2', 'key3', 'key4', 'key5']

newList = [dict(zip(fields, vals)) for vals in oldList]

answered Jan 20, 2012 at 20:20

g.d.d.c

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Comments

joaquin · Accepted Answer · 2012-01-20 20:43:29Z

2

If as you say you have a lot of entries, remember that python has namedtuples:

>>> tupleList = [('a1', 'b1', 'c1', 'd1', 'e1'),
...              ('a2', 'b2', 'c2', 'd2', 'e2'),
...              ('a10000', 'b10000', 'c10000', 'd10000', 'e100000')]
>>>
>>> from collections import namedtuple
>>> fv = namedtuple('fivevals', ('key1', 'key2', 'key3', 'key4', 'key5'))
>>> tuplelist = [fv(*item) for item in tupleList]
>>> 
>>> tuplelist[0].key1
'a1'
>>>

Namedtuples can be accesed by key but at the same time they are lightweight and require no more memory than regular tuples.

edited Jan 20, 2012 at 20:43

answered Jan 20, 2012 at 20:34

joaquin

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1 Comment

Kirk Strauser Over a year ago

Ack! I'm ashamed that I didn't remember that. Good call!

yak · Accepted Answer · 2012-01-20 20:21:58Z

0

from itertools import izip

keys = ['key1', 'key2', 'key3', 'key4', 'key5']
finalDictList = [dict(izip(names, x)) for x in tupleList]

answered Jan 20, 2012 at 20:21

yak

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2 Comments

yak Over a year ago

To avoid building a temporary list in each iteration of course.

Kirk Strauser Over a year ago

That seems like you'd be trading a tiny memory allocation for the overhead of a generator.

juliomalegria · Accepted Answer · 2012-01-20 20:32:04Z

0

finalDictList = []
for t in tupleList:
    finalDictList.append({
        'key1': t[0],
        'key2': t[1],
        'key3': t[2],
        'key4': t[3],
        'key5': t[4],
    })

Also, if that is possible, can we access each dictionary by it's key values, say, like finalDictList[0]['key1']?

Absolutely, that is exactly how you would do it.

edited Jan 20, 2012 at 20:32

juliomalegria

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answered Jan 20, 2012 at 20:20

Ian Clelland

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