Python 2.7 : delete item from list by value

Question

After performing some operations I get a list as following :

FreqItemset(items=[u'A_String_0'], freq=303)
FreqItemset(items=[u'A_String_0', u'Another_String_1'], freq=302)
FreqItemset(items=[u'B_String_1', u'A_String_0', u'A_OtherString_1'], freq=301)

I'd like to remove from list all items start from A_String_0 , but I'd like to keep other items (doesn't matter if A_String_0 exists in the middle or at the end of item )

So in example above delete lines 1 and 2 , keep line 3

I tried

 filter(lambda a: a != 'A_String_0', result)

and

result.remove('A_String_0')

all this doesn't help me

What do you mean by I'd like to remove from list all items start from A_String_0? — mbomb007
– mbomb007, Commented Dec 16, 2015 at 16:04
He wants to remove 'A_String_0' if it's the first element in the list, else leave it alone — wpercy
– wpercy, Commented Dec 16, 2015 at 16:04

zero323 · Accepted Answer · 2015-12-16 16:25:53Z

2

It is as simple as this:

from pyspark.mllib.fpm import FPGrowth

sets = [
    FPGrowth.FreqItemset(
       items=[u'A_String_0'], freq=303),
    FPGrowth.FreqItemset(
        items=[u'A_String_0', u'Another_String_1'], freq=302),
    FPGrowth.FreqItemset(
        items=[u'B_String_1', u'A_String_0', u'A_OtherString_1'], freq=301)
]

[x for x in sets if x.items[0] != 'A_String_0']
## [FreqItemset(items=['B_String_1', 'A_String_0', 'A_OtherString_1'], freq=301)]

In practice it would better to filter beffore collect:

filtered_sets = (model
    .freqItemsets()
    .filter(lambda x: x.items[0] != 'A_String_0')
    .collect())

edited Dec 16, 2015 at 16:25

answered Dec 16, 2015 at 16:04

zero323

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2 Comments

Toren Over a year ago

Can you please provide an example ? In case I'd like to search for 'A_S*' instead of 'A_String_0' ?

zero323 Over a year ago

x.items[0].startswith("A_S")

wpercy · Accepted Answer · 2015-12-16 16:04:39Z

2

How about result = result if result[0] != 'A_String_0' else result[1:]?

answered Dec 16, 2015 at 16:04

wpercy

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Comments

Baltasarq · Accepted Answer · 2015-12-16 16:09:22Z

2

It seems that you are using a list called FreqItemset. However, the name suggests that you should be using a set, instead of a list.

This way, you could have a set of searchable pairs string, frequency. For example:

>>> d = { "the": 2, "a": 3 }
>>> d[ "the" ]
2
>>> d[ "the" ] = 4
>>> d[ "a" ]
3
>>> del d[ "a" ]
>>> d
{'the': 4}

You can easily access each word (which is a key of the dictionary), change its value (its frequency of apparition), or remove it. All operations avoid the access to all the elements of the list, since it is a dictionary, i.e., its performance is good (better than using a list, anyway).

Just my two cents.

answered Dec 16, 2015 at 16:09

Baltasarq

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4 Comments

Toren Over a year ago

Thanks a lot for help . I'll try . About the type of Itemset , when I execute "print type (result) " I get a list . ( result = model....)

Baltasarq Over a year ago

Do you mean you cannot change it?

Toren Over a year ago

As I understand it's list of sets

Baltasarq Over a year ago

You should use the most appropriate data structure. If a list of sets does not suit you, then change it to a simple set.

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