Filtering a list based on another list

Question

I want to filter data based on items in drop.

data = [
    ['Basket', 'NBA ET', 'Kobe'],
    ['Basket', 'NCAA', 'Shaq'],
    ['Basket', 'ENG', 'Shaq'],
]
drop = ['NBA', 'NCAA']

And because I want the list with NBA ET to be left out too, it must be something beyond:

filtered = [d for d in data if d[1] not in drop]   # assume d[1] will hold

What I need is:

# pseudocode
filtered = [d for d in data if _ not in d[1] for _ in drop]

but I can never remember the syntax.

For the record, filtered should results in [['Basket', 'ENG', 'Shaq']]

Martijn Pieters · Accepted Answer · 2014-03-18 12:20:21Z

2

You can use any() and split the string on whitespace:

filtered = [d for d in data if not any(dropped in d[1].split() for dropped in drop)]

If you make drop a set, just test for an intersection:

drop = set(drop)

filtered = [d for d in data if not drop.intersection(d[1].split())]

The latter should be more performant the larger drop becomes.

Demo:

>>> data = [
...     ['Basket', 'NBA ET', 'Kobe'],
...     ['Basket', 'NCAA', 'Shaq'],
...     ['Basket', 'ENG', 'Shaq'],
... ]
>>> drop = ['NBA', 'NCAA']
>>> [d for d in data if not any(dropped in d[1].split() for dropped in drop)]
[['Basket', 'ENG', 'Shaq']]
>>> drop = set(drop)
>>> [d for d in data if not drop.intersection(d[1].split())]
[['Basket', 'ENG', 'Shaq']]

edited Mar 18, 2014 at 12:20

answered Mar 18, 2014 at 12:13

Martijn Pieters

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2 Comments

Martijn Pieters Over a year ago

@nutship: ick, yes, you want to remove not include. Corrected.

Martijn Pieters Over a year ago

@nutship: I wasn't downvoted; someone removed their upvote instead.

Matt · Accepted Answer · 2014-03-18 12:25:42Z

1

[row for row in data if not any(league in row[1].split() for league in drop)]
# [['Basket', 'ENG', 'Shaq']]

edited Mar 18, 2014 at 12:25

answered Mar 18, 2014 at 12:16

Matt

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6 Comments

Martijn Pieters Over a year ago

There is no need to create a list for any(). In fact, that defeats the purpose of using any() in the first place. Don't use a list comprehension here, use a generator expression instead (drop the [...] brackets from the any() call).

Matt Over a year ago

Strange, when I remove the list comprehension, the result is empty. What is the reason for this?

Martijn Pieters Over a year ago

No, your code works for me. Your code is identical to mine without the [..] (save for the names used).

Matt Over a year ago

I'm on python 2.7, could that be the reason?

Martijn Pieters Over a year ago

No, I use Python 2.7 too.

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