I have been for hours strugling to understand why i am not able to do this:
>>> import numpy as np >>> a = [np.empty((0,78,3)) for i in range(2)] >>> b = np.random.randint(10,size=(1,78,3)) >>> a[0] = np.append(a[0],[b],axis=0) Traceback (most recent call last): File "<stdin>", line 1, in <module> File "/usr/local/lib/python2.7/site-packages/numpy/lib/function_base.py", line 5003, in append return concatenate((arr, values), axis=axis) ValueError: all the input arrays must have same number of dimensions >>>
a is a list with s empty numpy arrays of shape (0,78,3)
b is a random numpy.array with shape (1,78,3)
I then try to append b to a[0]... but this doesn't seem to be possible because of not having same dimension?.. I am not sure whats the problem here.. if I removed the list part it would work, so why not with the list?..
Stay away from np.append. Learn to use np.concatenate correctly. This append just creates confusion.
Given your definitions, this works:
In [20]: a1 = [np.concatenate((i,b),axis=0) for i in a] In [21]: [i.shape for i in a1] Out[21]: [(1, 78, 3), (1, 78, 3)] In [22]: a Out[22]: [array([], shape=(0, 78, 3), dtype=float64), array([], shape=(0, 78, 3), dtype=float64)] In [23]: b.shape Out[23]: (1, 78, 3) In [24]: a1 = [np.concatenate((i,b),axis=0) for i in a] In [25]: [i.shape for i in a1] Out[25]: [(1, 78, 3), (1, 78, 3)]
A (0,78,3) can concatenate on axis 0 with a (1,78,3) array, producing another (1,78,3) array.
But why do it? It just makes a list with 2 copies of b.
c = [b,b]
does that just as well, and is simpler.
If you must collect many arrays of shape (78,3), do
alist = [] for _ in range(n): alist.append(np.ones((78,3)))
The resulting list of n arrays can be turned into an array with
np.array(alist) # (n, 78, 3) array
Or if you collect a list of (1,78,3) arrays, np.concatenate(alist, axis=0) will join them into the (n,78,3) array.
Your're not appending b but [b]. That doesn't work.
So in order to append b, use
a[0] = np.append(a[0],b,axis=0)
