CODEWITHSUNDEEP
pythonlistuniquedistinct-values
Hellnar
Hellnar

Reputation: 64793

Uniqueness for list of lists

I am curious what would be an efficient way of uniquifying such data objects:

testdata = [ ['9034968', 'ETH'], ['14160113', 'ETH'], ['9034968', 'ETH'], ['11111', 'NOT'], ['9555269', 'NOT'], ['15724032', 'ETH'], ['15481740', 'ETH'], ['15481757', 'ETH'], ['15481724', 'ETH'], ['10307528', 'ETH'], ['15481757', 'ETH'], ['15481724', 'ETH'], ['15481740', 'ETH'], ['15379365', 'ETH'], ['11111', 'NOT'], ['9555269', 'NOT'], ['15379365', 'ETH']
]

For each data pair, left numeric string PLUS the type at the right tells the uniqueness of a data element. The returned value should be a list of lists, the same as testdata, but only unique values should be kept.

Upvotes: 100

Views: 111815

Answers (8)

ZygD
ZygD

Reputation: 24366

Options for preserving order (Python 3.7+)

  • Inner lists become tuples:

    list(dict.fromkeys(map(tuple, testdata)))

    list({tuple(x): 1 for x in testdata})

  • Inner lists stay as lists (credits):

    list({tuple(x): x for x in testdata}.values())

    In case new list elements are a function of old ones, either a walrus operator := can be used

    list({tuple(y:=f(x)): y for x in testdata}.values())

    or we can turn inner lists into tuples and then back to lists

    list(map(list, {tuple(x): 1 for x in testdata}))

    list(map(list, dict.fromkeys(map(tuple, testdata))))

Upvotes: 3

Lou
Lou

Reputation: 2509

I was about to post my own take on this until I noticed that @pyfunc had already come up with something similar. I'll post my take on this problem anyway in case it's helpful.

testdata =[ ['9034968', 'ETH'], ['14160113', 'ETH'], ['9034968', 'ETH'], ['11111', 'NOT'], ['9555269', 'NOT'], ['15724032', 'ETH'], ['15481740', 'ETH'], ['15481757', 'ETH'], ['15481724', 'ETH'], ['10307528', 'ETH'], ['15481757', 'ETH'], ['15481724', 'ETH'], ['15481740', 'ETH'], ['15379365', 'ETH'], ['11111', 'NOT'], ['9555269', 'NOT'], ['15379365', 'ETH']
]
flatdata = [p[0] + "%" + p[1] for p in testdata]
flatdata = list(set(flatdata))
testdata = [p.split("%") for p in flatdata]
print(testdata)

Basically, you concatenate each element of your list into a single string using a list comprehension, so that you have a list of single strings. This is then much easier to turn into a set, which makes it unique. Then you simply split it on the other end and convert it back to your original list.

I don't know how this compares in terms of performance but it's a simple and easy-to-understand solution I think.

Upvotes: 1

Shaido
Shaido

Reputation: 28322

Use unique in numpy to solve this:

import numpy as np

np.unique(np.array(testdata), axis=0)

Note that the axis keyword needs to be specified otherwise the list is first flattened.

Alternatively, use vstack:

np.vstack({tuple(row) for row in testdata})

Upvotes: 6

Khan
Khan

Reputation: 1518

if you have a list of objects than you can modify @Mark Byers answer to:

unique_data = [list(x) for x in set(tuple(x.testList) for x in testdata)]

where testdata is a list of objects which has a list testList as attribute.

Upvotes: 1

Sam Morgan
Sam Morgan

Reputation: 3308

Expanding a bit on @Mark Byers solution, you can also just do one list comprehension and typecast to get what you need:

testdata = list(set(tuple(x) for x in testdata))

Also, if you don't like list comprehensions as many find them confusing, you can do the same in a for loop:

for i, e in enumerate(testdata):
    testdata[i] = tuple(e)
testdata = list(set(testdata))

Upvotes: 3

Manoj Govindan
Manoj Govindan

Reputation: 74695

I tried @Mark's answer and got an error. Converting the list and each elements into a tuple made it work. Not sure if this the best way though.

list(map(list, set(map(lambda i: tuple(i), testdata))))

Of course the same thing can be expressed using a list comprehension instead.

[list(i) for i in set(tuple(i) for i in testdata)]

I am using Python 2.6.2.

Update

@Mark has since changed his answer. His current answer uses tuples and will work. So will mine :)

Update 2

Thanks to @Mark. I have changed my answer to return a list of lists rather than a list of tuples.

Upvotes: 11

pyfunc
pyfunc

Reputation: 66709

import sets
testdata =[ ['9034968', 'ETH'], ['14160113', 'ETH'], ['9034968', 'ETH'], ['11111', 'NOT'], ['9555269', 'NOT'], ['15724032', 'ETH'], ['15481740', 'ETH'], ['15481757', 'ETH'], ['15481724', 'ETH'], ['10307528', 'ETH'], ['15481757', 'ETH'], ['15481724', 'ETH'], ['15481740', 'ETH'], ['15379365', 'ETH'], ['11111', 'NOT'], ['9555269', 'NOT'], ['15379365', 'ETH']]
conacatData = [x[0] + x[1] for x in testdata]
print conacatData
uniqueSet = sets.Set(conacatData)
uniqueList = [ [t[0:-3], t[-3:]] for t in uniqueSet]
print uniqueList

Upvotes: 1

Mark Byers
Mark Byers

Reputation: 838056

You can use a set:

unique_data = [list(x) for x in set(tuple(x) for x in testdata)]

You can also see this page which benchmarks a variety of methods that either preserve or don't preserve order.

Upvotes: 173

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