testing whether a Numpy array contains a given row
You can use .tolist()
>>> a = np.array([[1,2],[10,20],[100,200]])
>>> [1,2] in a.tolist()
True
>>> [1,20] in a.tolist()
False
>>> [1,20] in a.tolist()
False
>>> [1,42] in a.tolist()
False
>>> [42,1] in a.tolist()
False
Or use a view:
>>> any((a[:]==[1,2]).all(1))
True
>>> any((a[:]==[1,20]).all(1))
False
Or generate over the numpy list (potentially VERY SLOW):
any(([1,2] == x).all() for x in a) # stops on first occurrence
Or use numpy logic functions:
any(np.equal(a,[1,2]).all(1))
If you time these:
import numpy as np
import time
n=300000
a=np.arange(n*3).reshape(n,3)
b=a.tolist()
t1,t2,t3=a[n//100][0],a[n//2][0],a[-10][0]
tests=[ ('early hit',[t1, t1+1, t1+2]),
('middle hit',[t2,t2+1,t2+2]),
('late hit', [t3,t3+1,t3+2]),
('miss',[0,2,0])]
fmt='\t{:20}{:.5f} seconds and is {}'
for test, tgt in tests:
print('\n{}: {} in {:,} elements:'.format(test,tgt,n))
name='view'
t1=time.time()
result=(a[...]==tgt).all(1).any()
t2=time.time()
print(fmt.format(name,t2-t1,result))
name='python list'
t1=time.time()
result = True if tgt in b else False
t2=time.time()
print(fmt.format(name,t2-t1,result))
name='gen over numpy'
t1=time.time()
result=any((tgt == x).all() for x in a)
t2=time.time()
print(fmt.format(name,t2-t1,result))
name='logic equal'
t1=time.time()
np.equal(a,tgt).all(1).any()
t2=time.time()
print(fmt.format(name,t2-t1,result))
You can see that hit or miss, the numpy routines are the same speed to search the array. The Python in operator is potentially a lot faster for an early hit, and the generator is just bad news if you have to go all the way through the array.
Here are the results for 300,000 x 3 element array:
early hit: [9000, 9001, 9002] in 300,000 elements:
view 0.01002 seconds and is True
python list 0.00305 seconds and is True
gen over numpy 0.06470 seconds and is True
logic equal 0.00909 seconds and is True
middle hit: [450000, 450001, 450002] in 300,000 elements:
view 0.00915 seconds and is True
python list 0.15458 seconds and is True
gen over numpy 3.24386 seconds and is True
logic equal 0.00937 seconds and is True
late hit: [899970, 899971, 899972] in 300,000 elements:
view 0.00936 seconds and is True
python list 0.30604 seconds and is True
gen over numpy 6.47660 seconds and is True
logic equal 0.00965 seconds and is True
miss: [0, 2, 0] in 300,000 elements:
view 0.00936 seconds and is False
python list 0.01287 seconds and is False
gen over numpy 6.49190 seconds and is False
logic equal 0.00965 seconds and is False
And for 3,000,000 x 3 array:
early hit: [90000, 90001, 90002] in 3,000,000 elements:
view 0.10128 seconds and is True
python list 0.02982 seconds and is True
gen over numpy 0.66057 seconds and is True
logic equal 0.09128 seconds and is True
middle hit: [4500000, 4500001, 4500002] in 3,000,000 elements:
view 0.09331 seconds and is True
python list 1.48180 seconds and is True
gen over numpy 32.69874 seconds and is True
logic equal 0.09438 seconds and is True
late hit: [8999970, 8999971, 8999972] in 3,000,000 elements:
view 0.09868 seconds and is True
python list 3.01236 seconds and is True
gen over numpy 65.15087 seconds and is True
logic equal 0.09591 seconds and is True
miss: [0, 2, 0] in 3,000,000 elements:
view 0.09588 seconds and is False
python list 0.12904 seconds and is False
gen over numpy 64.46789 seconds and is False
logic equal 0.09671 seconds and is False
Which seems to indicate that np.equal is the fastest pure numpy way to do this...
Numpys __contains__ is, at the time of writing this, (a == b).any() which is arguably only correct if b is a scalar (it is a bit hairy, but I believe – works like this only in 1.7. or later – this would be the right general method (a == b).all(np.arange(a.ndim - b.ndim, a.ndim)).any(), which makes sense for all combinations of a and b dimensionality)...
EDIT: Just to be clear, this is not necessarily the expected result when broadcasting is involved. Also someone might argue that it should handle the items in a separately as np.in1d does. I am not sure there is one clear way it should work.
Now you want numpy to stop when it finds the first occurrence. This AFAIK does not exist at this time. It is difficult because numpy is based mostly on ufuncs, which do the same thing over the whole array.
Numpy does optimize these kind of reductions, but effectively that only works when the array being reduced is already a boolean array (i.e. np.ones(10, dtype=bool).any()).
Otherwise it would need a special function for __contains__ which does not exist. That may seem odd, but you have to remember that numpy supports many data types and has a bigger machinery to select the correct ones and select the correct function to work on it. So in other words, the ufunc machinery cannot do it, and implementing __contains__ or such specially is not actually that trivial because of data types.
You can of course write it in python, or since you probably know your data type, writing it yourself in Cython/C is very simple.
That said. Often it is much better anyway to use sorting based approach for these things. That is a little tedious as well as there is no such thing as searchsorted for a lexsort, but it works (you could also abuse scipy.spatial.cKDTree if you like). This assumes you want to compare along the last axis only:
# Unfortunatly you need to use structured arrays:
sorted = np.ascontiguousarray(a).view([('', a.dtype)] * a.shape[-1]).ravel()
# Actually at this point, you can also use np.in1d, if you already have many b
# then that is even better.
sorted.sort()
b_comp = np.ascontiguousarray(b).view(sorted.dtype)
ind = sorted.searchsorted(b_comp)
result = sorted[ind] == b_comp
This works also for an array b, and if you keep the sorted array around, is also much better if you do it for a single value (row) in b at a time, when a stays the same (otherwise I would just np.in1d after viewing it as a recarray). Important: you must do the np.ascontiguousarray for safety. It will typically do nothing, but if it does, it would be a big potential bug otherwise.
I think
equal([1,2], a).all(axis=1) # also, ([1,2]==a).all(axis=1)
# array([ True, False, False], dtype=bool)
will list the rows that match. As Jamie points out, to know whether at least one such row exists, use any:
equal([1,2], a).all(axis=1).any()
# True
Aside:
I suspect in (and __contains__) is just as above but using any instead of all.
I've compared the suggested solutions with perfplot and found that, if you're looking for a 2-tuple in a long unsorted list,
np.any(np.all(a == b, axis=1))
is the fastest solution. An explicit short-circuit loop can always be faster if a match is found in the first few rows.
Code to reproduce the plot:
import numpy as np
import perfplot
target = [6, 23]
def setup(n):
return np.random.randint(0, 100, (n, 2))
def any_all(data):
return np.any(np.all(target == data, axis=1))
def tolist(data):
return target in data.tolist()
def loop(data):
for row in data:
if np.all(row == target):
return True
return False
def searchsorted(a):
s = np.ascontiguousarray(a).view([('', a.dtype)] * a.shape[-1]).ravel()
s.sort()
t = np.ascontiguousarray(target).view(s.dtype)
ind = s.searchsorted(t)
return (s[ind] == t)[0]
perfplot.save(
"out02.png",
setup=setup,
kernels=[any_all, tolist, loop, searchsorted],
n_range=[2 ** k for k in range(2, 20)],
xlabel="len(array)",
)