Finding intersection/difference between python lists

Question

I have two python lists:

a = [(\'when\', 3), (\'why\', 4), (\'throw\', 9), (\'send\', 15), (\'you\', 1)]

b = [\'the\', \'when\', \'send\', \'we\', \'us\']
<         


        

Answer 1

A list comprehension will work.

a = [('when', 3), ('why', 4), ('throw', 9), ('send', 15), ('you', 1)]
b = ['the', 'when', 'send', 'we', 'us']
filtered = [i for i in a if not i[0] in b]

>>>print(filtered)
[('why', 4), ('throw', 9), ('you', 1)]

Answer 2

As this is tagged with numpy, here is a numpy solution using numpy.in1d benchmarked against the list comprehension:

In [1]: a = [('when', 3), ('why', 4), ('throw', 9), ('send', 15), ('you', 1)]

In [2]: b = ['the', 'when', 'send', 'we', 'us']

In [3]: a_ar = np.array(a, dtype=[('string','|S5'), ('number',float)])

In [4]: b_ar = np.array(b)

In [5]: %timeit filtered = [i for i in a if not i[0] in b]
1000000 loops, best of 3: 778 ns per loop

In [6]: %timeit filtered = a_ar[-np.in1d(a_ar['string'], b_ar)]
10000 loops, best of 3: 31.4 us per loop

So for 5 records the list comprehension is faster.

However for large data sets the numpy solution is twice as fast as the list comprehension:

In [7]: a = a * 1000

In [8]: a_ar = np.array(a, dtype=[('string','|S5'), ('number',float)])

In [9]: %timeit filtered = [i for i in a if not i[0] in b]
1000 loops, best of 3: 647 us per loop

In [10]: %timeit filtered = a_ar[-np.in1d(a_ar['string'], b_ar)]
1000 loops, best of 3: 302 us per loop

Answer 3

Use filter:

c = filter(lambda (x, y): False if x in b else True, a)

Answer 4

Easy way

a = [('when', 3), ('why', 4), ('throw', 9), ('send', 15), ('you', 1)]
b = ['the', 'when', 'send', 'we', 'us']
c=[] # a list to store the required tuples 
#compare the first element of each tuple in with an element in b
for i in a:
    if i[0] not in b:
        c.append(i)
print(c)

Answer 5

Try this :

a = [('when', 3), ('why', 4), ('throw', 9), ('send', 15), ('you', 1)]

b = ['the', 'when', 'send', 'we', 'us']

c=[]

for x in a:
    if x[0] not in b:
        c.append(x)
print c

Demo: http://ideone.com/zW7mzY

Answer 6

A list comprehension should work:

c = [item for item in a if item[0] not in b]

Or with a dictionary comprehension:

d = dict(a)
c = {key: value for key in d.iteritems() if key not in b}