Input
array([[10, 11, 12, 13, 26, 28, 11],
[ 4, 9, 19, 20, 21, 26, 12],
[ 1, 7, 10, 23, 28, 32, 16],
[ 4, 6, 7, 10, 13, 25, 3],
[ 4, 6, 15, 17, 30, 31, 16]])
test = np.array([10, 11, 12, 13, 26, 28, 11])
Trying
def check(test,array):
for i in array:
if np.array_equal(i,test):
print('in')
break
print('not in')
Is there any simple method without using for...in...if...else structure?
There are indeed ways. The most pythonic one is probably to use a generator comprehension.
def check(test,array):
return any(numpy.array_equal(x, test) for x in array)
Note that this is not quite the same behaviour as it returns a Boolean rather than printing the result, but that behaviour should be easy enough to add outside.
As an aside, I don't like the variable name test. It sounds to me like the name of a predicate function rather than the name of an array you want to look for.
The name check is also rather too general to usefully say what it is checking.
Now I'm perhaps overly fond of slightly too descriptive names, but I'd probably be calling this def is_row_in_array(row, array):
any(map(lambda x:np.array_equal(x,test), array)) ? Is it more pythonic?
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If you don't like the looping and the if ... statements you could use solely NumPy methods:
def is_row_in_array(row , arr):
return np.any(np.sum(np.abs(arr-row), axis=1) == 0)
(arr == row).all(axis=1).any() is shorter and a bit faster, but the general approach in this answer is right.
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