Remove tuple from list of tuples if certain condition is met

Question

I have a list of tuples that look like this;

ListTuples = [(100, 'AAA'), (80, 'BBB'), (20, 'CCC'), (40, 'DDD')]

I want to remove the tuples when the first element of the tuple is less than 50. The OutputList will look like this;

OutputList = [(100, 'AAA'), (80, 'BBB')]

How can this be done in python?

Thank you very much for your help.

Tobit

Answer 1

You can easily do it as:

out_tup = [i for i in in_tup if i[0] >= 50]

[Out]: [(100, 'AAA'), (80, 'BBB')]

This simply creates a new list of tuples with only those tuples whose first element is greater than or equal to 50. Same result, however the approach is different. Instead of removing invalid tuples you accept the valid ones.

Grosmark

Answer 2

You can also do:

>>> OutputList = filter(ListTuples, lambda x: x[0] >= 50)
>>> OutputList
[(100, 'AAA'), (80, 'BBB')]

Chengtu

May I ask whether using filter and lambda be slower than the answer provided by sshashank124? – Tobit

@user3293156 generally filter is slightly faster as it is a builtin function. – Chengtu

@AlexThornton Only when used with built-in functions, with lambdas it is slow. – Safier

@Aशwiniचhaudhary I suppose, but the difference should be marginal. – Chengtu

Wow! But due to my incompetence in python, this code using filter and lambda is less readable compared to sshashank124's. I have to study what these 2 functions are about first. – Tobit

@ Aशwini चhaudhary: Do you mean code with lambdas are slower? – Tobit

Note that this raises an error. The syntax needs to be filter(function, list), so OutputList = filter(lambda x: x[0] >= 50, ListTuples ) – Illaffected

Answer 3

Code snippet for timing the solutions given by sshashank124 and Alex Thornton:

ListTuples = [(100, 'AAA'), (80, 'BBB'), (20, 'CCC'), (40, 'DDD')]
import timeit
timeit.timeit('[i for i in ListTuples if i[0] >= 50]', globals=globals(), number=1000000)
>>> 0.6041104920150246

timeit.timeit('filter(lambda x: x[0] >= 50, ListTuples)', globals=globals(), number=1000000)
>>> 0.3009479799948167

The build-in filter() solution is faster for this example.

Sunstone

Answer 4

We can Do this with simple counter and a new list:

   new = []

   for i in ListTuples:
        for j in i:
            if ( counter % 2 == 0 ):
                if ( j > 50):
                    new.append(i)
            counter = counter +1

output :

[(100, 'AAA') , (80, 'BBB')]

Godolphin

Answer 5

Try this,

>>> ListTuples = [(100, 'AAA'), (80, 'BBB'), (20, 'CCC'), (40, 'DDD')]
>>> new=[]
>>> for i in ListTuples:
    if i[0]>50:
        new.append(i)


>>> new
[(100, 'AAA'), (80, 'BBB')]
>>>

Cigar

This code looks familiar. It looks like the C language kind of syntax. But is the execution slower than the pythonic code provided by Alex Thornton and sshashank124? – Tobit

Answer 6

ListTuples = [(100, 'AAA'), (80, 'BBB'), (20, 'CCC'), (40, 'DDD')]
out=[]
for item in ListTuples:
    if item[0]>50:
        out.append((item[0],item[1]))
print(out)

Hofuf