查找python列表之间的交集/差异

Question

I have two python lists:

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

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

I need to filter out all the elements from a that are similar to those in b. Like in this case, I should get:

c = [('why', 4), ('throw', 9), ('you', 1)]

What should be the most effective way?

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

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}

Answer 3

in is nice, but you should use sets at least for b . If you have numpy, you could also try np.in1d of course, but if it is faster or not, you should probably try.

# ruthless copy, but use the set...
b = set(b)
filtered = [i for i in a if not i[0] in b]

# with numpy (note if you create the array like this, you must already put
# the maximum string length, here 10), otherwise, just use an object array.
# its slower (likely not worth it), but safe.
a = np.array(a, dtype=[('key', 's10'), ('val', int)])
b = np.asarray(b)

mask = ~np.in1d(a['key'], b)
filtered = a[mask]

Sets also have have the methods difference , etc. which probably are not to useful here, but in general probably are.

Answer 4

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 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

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 7

使用过滤器：

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