I have a numpy matrix and would like to concatenate all of the rows together so I end up with one long array.
#example
input:
[[1 2 3]
[4 5 6}
[7 8 9]]
output:
[[1 2 3 4 5 6 7 8 9]]
The way I am doing it now doe not seem pythonic. I'm sure there is a better way.
combined_x = x[0]
for index, row in enumerate(x):
if index!= 0:
combined_x = np.concatenate((combined_x,x[index]),axis=1)
Thank you for the help.
concat function is 50 times faster than using the DataFrame. append version. With multiple append , a new DataFrame is created at each iteration, and the underlying data is copied each time.
Numpy arrays are mutable objects that have clearly defined in place operations. If a and b are arrays of the same shape, a += b adds the two arrays together, using a as an output buffer.
I would suggest the ravel
or flatten
method of ndarray
.
>>> a = numpy.arange(9).reshape(3, 3)
>>> a.ravel()
array([0, 1, 2, 3, 4, 5, 6, 7, 8])
ravel
is faster than concatenate
and flatten
because it doesn't return a copy unless it has to:
>>> a.ravel()[5] = 99
>>> a
array([[ 0, 1, 2],
[ 3, 4, 99],
[ 6, 7, 8]])
>>> a.flatten()[5] = 77
>>> a
array([[ 0, 1, 2],
[ 3, 4, 99],
[ 6, 7, 8]])
But if you need a copy to avoid the memory sharing illustrated above, you're better off using flatten
than concatenate
, as you can see from these timings:
>>> %timeit a.ravel()
1000000 loops, best of 3: 468 ns per loop
>>> %timeit a.flatten()
1000000 loops, best of 3: 1.42 us per loop
>>> %timeit numpy.concatenate(a)
100000 loops, best of 3: 2.26 us per loop
Note also that you can achieve the exact result that your output illustrates (a one-row 2-d array) with reshape
(thanks Pierre GM!):
>>> a = numpy.arange(9).reshape(3, 3)
>>> a.reshape(1, -1)
array([[0, 1, 2, 3, 4, 5, 6, 7, 8]])
>>> %timeit a.reshape(1, -1)
1000000 loops, best of 3: 736 ns per loop
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