python sort dictionary by value array

Question

I have a dictionary with an array as elements. Say:

masterListShort = {'a': [5, 2, 1, 2], 'b': [7, 2, 4, 1], 'c': [2, 0, 1, 1]}

I would like to reverse sort this dictionary by the first element of the values. I would then like to write my output to a tab delimited file like this:

<key>    <value1>    <value2>    <value3>    etc.

My current code where I write my dictionary to file looks like this:

# write the masterListShort to file
outFile2 = open('../masterListShort.tsv', 'w')
for item in sorted(masterListShort):
    tempStr = '\t'.join(map(str, masterListShort[item]))
    outFile2.write(str(item) + '\t' + tempStr + '\n')

outFile2.close()

This code works fine, it just does not sort the list. I want my output to be written in a tab delimited file format. So:

b    7    2    4    1
c    5    2    1    2
a    2    0    1    1

I have found the following commands so far, and was wondering if i could apply them to my code:

import operator
sorted(myDict, key=operator.itemgetter(1))

Please do test your samples; your sample dictionary is using entirely incorrect syntax. — Martijn Pieters
– Martijn Pieters, Commented May 7, 2014 at 10:52

Martijn Pieters · Accepted Answer · 2014-05-07 12:01:29Z

8

You'll need to sort the values then, on the first index (so 0 for zero-based indexing), and tell sorted() to reverse the order:

import operator

sorted(myDict.values(), key=operator.itemgetter(0), reverse=True)

Without the dict.values() call you are trying to sort the keys instead.

Demo:

>>> import operator
>>> myDict = {'a': [5, 2, 1, 2], 'b': [7, 2, 4, 1], 'c': [2, 0, 1, 1]}
>>> sorted(myDict.values(), key=operator.itemgetter(0), reverse=True)
[[7, 2, 4, 1], [5, 2, 1, 2], [2, 0, 1, 1]]

If you wanted to output key-value pairs, then use dict.items() and use lambda to access the first index on just the value:

sorted(myDict.items(), key=lambda i: i[1][0], reverse=True)

Demo:

>>> sorted(myDict.items(), key=lambda i: i[1][0], reverse=True)
[('b', [7, 2, 4, 1]), ('a', [5, 2, 1, 2]), ('c', [2, 0, 1, 1])]

In both cases, there is actually not that much point on sorting just by the first element; you can just leave the lists to sort naturally:

sorted(myDict.values(), reverse=True)  # sort just the values
sorted(myDict.items(), key=operator.itemgetter(1), reverse=True)  # sort items

as a list of lists is sorted in lexicographical ordering; if the first elements are equal, then two lists are ordered based on the second element, etc.

To write this out to a tab-delimited file, use the csv module; it'll take care of converting values to strings, writing the tab delimiters and handle newlines:

import csv

with open('../masterListShort.tsv', 'wb') as outfh:
    writer = csv.writer(outfh, delimiter='\t')
    for key, values in sorted(myDict.items(), key=operator.itemgetter(1), reverse=True):
        writer.writerow([key] + values)

edited May 7, 2014 at 12:01

answered May 7, 2014 at 10:53

Martijn Pieters

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4 Comments

MrFronk Over a year ago

Yes, I had this. the problem I have here is that I cannot get the original key anymore. (I would like to write my output as key + value)

Martijn Pieters Over a year ago

@MrFronk: Then why did you not ask for that instead? Your sample output showed just the values.

MrFronk Over a year ago

I reformulated my question, I hope the quetion is clear now. Sorry for inconvenience caused

Martijn Pieters Over a year ago

@MrFronk: I've added a section on writing the TSV file for you too now; it'll write key value0 value1 ... valueN per row, with the rows sorted on value0.

will · Accepted Answer · 2014-05-07 12:04:36Z

0

You can sort the list of tuples that dict.items() gives;

myDict = {'a': [5, 2, 1, 2], 'b': [7, 2, 4, 1], 'c': [2, 0, 1, 1]}

print myDict.items()
#[('a', [5, 2, 1, 2]), ('c', [2, 0, 1, 1]), ('b', [7, 2, 4, 1])]

print sorted(myDict.items(), key=lambda x: x[1][0], reverse=True))
#[('c', [2, 0, 1, 1]), ('a', [5, 2, 1, 2]), ('b', [7, 2, 4, 1])]

the key keyword argument of sorted is a function you provide it with, which returns something the function can use to sort each element of the list you give it to be sorted on.

lambda x: x[1][0] is just a quick way of making a function which you could have used instead;

i.e.

print sorted(myDict.items(), key=lambda x: x[1][0])

is the same as

def sortFunction(x):
  return x[1][0]

print sorted(myDict.items(), key=sortFunction)

The reason it's giving x[1][0] is because the list myDict.items() is a 2D list; each element is a tuple, where the first element is the key, and the second element is the value. so x[1] is the value (i.e. the lists assigned to each key). And you want to sort on their first values, so we sort on x[1][0].

Now it's just a case of printing them in the format you want;

for key, value in sorted(myDict.items(), key=sortFunction, reverse=True):
  print key + "\t" + "\t".join([str(v) for v in value])

That will give:

b   7   2   4   1
a   5   2   1   2
c   2   0   1   1

edited May 7, 2014 at 12:04

answered May 7, 2014 at 11:23

will

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1 Comment

MrFronk Over a year ago

I reformulated my question, I hope the quetion is clear now. Sorry for inconvenience caused

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