I am trying to fill an empty(not np.empty!) array with values using append but I am gettin error:
My code is as follows:
import numpy as np
result=np.asarray([np.asarray([]),np.asarray([])])
result[0]=np.append([result[0]],[1,2])
And I am getting:
ValueError: could not broadcast input array from shape (2) into shape (0)
I might understand the question incorrectly, but if you want to declare an array of a certain shape but with nothing inside, the following might be helpful:
Initialise empty array:
>>> a = np.zeros((0,3)) #or np.empty((0,3)) or np.array([]).reshape(0,3)
>>> a
array([], shape=(0, 3), dtype=float64)
Now you can use this array to append rows of similar shape to it. Remember that a numpy array is immutable, so a new array is created for each iteration:
>>> for i in range(3):
... a = np.vstack([a, [i,i,i]])
...
>>> a
array([[ 0., 0., 0.],
[ 1., 1., 1.],
[ 2., 2., 2.]])
np.vstack and np.hstack is the most common method for combining numpy arrays, but coming from Matlab I prefer np.r_ and np.c_:
Concatenate 1d:
>>> a = np.zeros(0)
>>> for i in range(3):
... a = np.r_[a, [i, i, i]]
...
>>> a
array([ 0., 0., 0., 1., 1., 1., 2., 2., 2.])
Concatenate rows:
>>> a = np.zeros((0,3))
>>> for i in range(3):
... a = np.r_[a, [[i,i,i]]]
...
>>> a
array([[ 0., 0., 0.],
[ 1., 1., 1.],
[ 2., 2., 2.]])
Concatenate columns:
>>> a = np.zeros((3,0))
>>> for i in range(3):
... a = np.c_[a, [[i],[i],[i]]]
...
>>> a
array([[ 0., 1., 2.],
[ 0., 1., 2.],
[ 0., 1., 2.]])
numpy.append is pretty different from list.append in python. I know that's thrown off a few programers new to numpy. numpy.append is more like concatenate, it makes a new array and fills it with the values from the old array and the new value(s) to be appended. For example:
import numpy
old = numpy.array([1, 2, 3, 4])
new = numpy.append(old, 5)
print old
# [1, 2, 3, 4]
print new
# [1, 2, 3, 4, 5]
new = numpy.append(new, [6, 7])
print new
# [1, 2, 3, 4, 5, 6, 7]
I think you might be able to achieve your goal by doing something like:
result = numpy.zeros((10,))
result[0:2] = [1, 2]
# Or
result = numpy.zeros((10, 2))
result[0, :] = [1, 2]
Update:
If you need to create a numpy array using loop, and you don't know ahead of time what the final size of the array will be, you can do something like:
import numpy as np
a = np.array([0., 1.])
b = np.array([2., 3.])
temp = []
while True:
rnd = random.randint(0, 100)
if rnd > 50:
temp.append(a)
else:
temp.append(b)
if rnd == 0:
break
result = np.array(temp)
In my example result will be an (N, 2) array, where N is the number of times the loop ran, but obviously you can adjust it to your needs.
new update
The error you're seeing has nothing to do with types, it has to do with the shape of the numpy arrays you're trying to concatenate. If you do np.append(a, b) the shapes of a and b need to match. If you append an (2, n) and (n,) you'll get a (3, n) array. Your code is trying to append a (1, 0) to a (2,). Those shapes don't match so you get an error.
This error arise from the fact that you are trying to define an object of shape (0,) as an object of shape (2,). If you append what you want without forcing it to be equal to result[0] there is no any issue:
b = np.append([result[0]], [1,2])
But when you define result[0] = b you are equating objects of different shapes, and you can not do this. What are you trying to do?
Here's the result of running your code in Ipython. Note that result is a (2,0) array, 2 rows, 0 columns, 0 elements. The append produces a (2,) array. result[0] is (0,) array. Your error message has to do with trying to assign that 2 item array into a size 0 slot. Since result is dtype=float64, only scalars can be assigned to its elements.
In [65]: result=np.asarray([np.asarray([]),np.asarray([])])
In [66]: result
Out[66]: array([], shape=(2, 0), dtype=float64)
In [67]: result[0]
Out[67]: array([], dtype=float64)
In [68]: np.append(result[0],[1,2])
Out[68]: array([ 1., 2.])
np.array is not a Python list. All elements of an array are the same type (as specified by the dtype). Notice also that result is not an array of arrays.
Result could also have been built as
ll = [[],[]]
result = np.array(ll)
while
ll[0] = [1,2]
# ll = [[1,2],[]]
the same is not true for result.
np.zeros((2,0)) also produces your result.
Actually there's another quirk to result.
result[0] = 1
does not change the values of result. It accepts the assignment, but since it has 0 columns, there is no place to put the 1. This assignment would work in result was created as np.zeros((2,1)). But that still can't accept a list.
But if result has 2 columns, then you can assign a 2 element list to one of its rows.
result = np.zeros((2,2))
result[0] # == [0,0]
result[0] = [1,2]
What exactly do you want result to look like after the append operation?
numpy.append always copies the array before appending the new values. Your code is equivalent to the following:
import numpy as np
result = np.zeros((2,0))
new_result = np.append([result[0]],[1,2])
result[0] = new_result # ERROR: has shape (2,0), new_result has shape (2,)
Perhaps you mean to do this?
import numpy as np
result = np.zeros((2,0))
result = np.append([result[0]],[1,2])
SO thread 'Multiply two arrays element wise, where one of the arrays has arrays as elements' has an example of constructing an array from arrays. If the subarrays are the same size, numpy makes a 2d array. But if they differ in length, it makes an array with dtype=object, and the subarrays retain their identity.
Following that, you could do something like this:
In [5]: result=np.array([np.zeros((1)),np.zeros((2))])
In [6]: result
Out[6]: array([array([ 0.]), array([ 0., 0.])], dtype=object)
In [7]: np.append([result[0]],[1,2])
Out[7]: array([ 0., 1., 2.])
In [8]: result[0]
Out[8]: array([ 0.])
In [9]: result[0]=np.append([result[0]],[1,2])
In [10]: result
Out[10]: array([array([ 0., 1., 2.]), array([ 0., 0.])], dtype=object)
However, I don't offhand see what advantages this has over a pure Python list or lists. It does not work like a 2d array. For example I have to use result[0][1], not result[0,1]. If the subarrays are all the same length, I have to use np.array(result.tolist()) to produce a 2d array.
Related
I made a numpy array of size 6*6*51 with the command
matrix = np.zeros((6,6,51))
But when I write a loop so that a 6 element array is placed in matrix[i][:][j] it says the
ValueError: could not broadcast input array from shape (6) into shape
(51)
What am I doing wrong here ?
import numpy as np
a = np.zeros((6,6,51))
You are indexing incorrectly. Your subscription should include all the dimensions.
a[i][:] is essentially the same as a[i] then j in a[i][:][j] is indexing the second dimension.
In [14]: b = a[1][:]
In [15]: b.shape
Out[15]: (6, 51)
In [16]: c = a[1][:][1]
In [17]: c.shape
Out[17]: (51,)
Use a[i,:,j] for the subscription.
In [20]: a[1,:,1] = np.arange(6)
In [21]: a[1,:,1]
Out[21]: array([ 0., 1., 2., 3., 4., 5.])
Numpy Indexing
Is there a way to add (as opposed to sum) multiple arrays together in a single operation? Obviously, np.sum and np.add are different operations, however, the problem I'm struggling with right now is that np.add only takes two arrays at once. I could utilize either
output = 0
for arr in arr_list:
output = output + array
or
output = 0
for arr in arr_list:
output = np.add(output, array)
and, yes, this is workable. However, it would be nice if I could simply do some variant of
output = np.add_multiple(arr_list)
Does this exist?
EDIT:
I failed to be clear initially. I specifically require a function that does not require an array of arrays to be constructed, as my arrays are not of equal dimensions and require broadcasting, for example:
a = np.arange(3).reshape(1,3)
b = np.arange(9).reshape(3,3)
a, b = a[:,:,None,None], b[None,None,:,:]
These work:
a + b # Works
np.add(a, b) # Works
These do not, and fail with the same exception:
np.sum([a, b], axis = 0)
np.add.reduce([a, b])
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ValueError: could not broadcast input array from shape (3,1,1) into shape (1)
You can just use the Python's sum built-in:
output = sum(arr_list)
For many other numpy functions, np.<ufunc>.reduce can be used as shown by #hpaulj.
You can use the sum() to add multiple arrays.
arr = np.array([[6,2,3,5,4,3],
[7,7,2,4,6,7],
[10,6,2,4,5,9]])
np.add(0, arr.sum(axis=0))
In [18]: alist = [np.arange(4),np.ones(4),np.array([10,1,11,2])]
In [19]: np.add.reduce(alist)
Out[19]: array([11., 3., 14., 6.])
In [20]: alist[0]+alist[1]+alist[2]
Out[20]: array([11., 3., 14., 6.])
And for more fun:
In [21]: np.add.accumulate(alist)
Out[21]:
array([[ 0., 1., 2., 3.],
[ 1., 2., 3., 4.],
[11., 3., 14., 6.]])
edit
In [53]: a.shape
Out[53]: (1, 1, 1, 3)
In [54]: b.shape
Out[54]: (3, 3, 1, 1)
Addition with broadcasting:
In [63]: sum([a,b]).shape
Out[63]: (3, 3, 1, 3)
In [64]: (a+b).shape
Out[64]: (3, 3, 1, 3)
In [66]: np.add.reduce([a,b]).shape
Out[66]: (3, 3, 1, 3)
For what it's worth, I was suggesting add.reduce because I thought you wanted to add more than 2 arrays.
All these work as long as the arrays broadcast together.
I want to combine two arrays of different sizes into a third array(which is an array of arrays).
I have tried to use np.concatenate function as well as the np.append function but am not getting the desired results.
A=[1. 1. 0.]
B=[0. 1.]
y2=np.concatenate((yl, yr))
Expected Result [[0 1],[1 1 0]]
What are you trying to do with the result?
Just use a list if you want to iterate over the array elements.
import numpy as np
A=np.array([1., 1., 0.])
B=np.array([0., 1.])
y2=[A,B]
You can use np.array here.
import numpy as np
A = np.array([1., 1., 0.])
B = np.array([0., 1.])
y2 = np.array([A,B])
print(y2)
#output:- array([array([1., 1., 0.]), array([0., 1.])])
And for your expected output, you need to convert A and B to list:-
v = np.vectorize(int)
y2 = [v(B).tolist(), v(A).tolist()]
print(y2)
#output:- [[0, 1], [1, 1, 0]]
I have the following question. Is there somekind of method with numpy or scipy , which I can use to get an given unsorted array like this
a = np.array([0,0,1,1,4,4,4,4,5,1891,7]) #could be any number here
to something where the numbers are interpolated/mapped , there is no gap between the values and they are in the same order like before?:
[0,0,1,1,2,2,2,2,3,5,4]
EDIT
Is it furthermore possible to swap/shuffle the numbers after the mapping, so that
[0,0,1,1,2,2,2,2,3,5,4]
become something like:
[0,0,3,3,5,5,5,5,4,1,2]
Edit: I'm not sure what the etiquette is here (should this be a separate answer?), but this is actually directly obtainable from np.unique.
>>> u, indices = np.unique(a, return_inverse=True)
>>> indices
array([0, 0, 1, 1, 2, 2, 2, 2, 3, 5, 4])
Original answer: This isn't too hard to do in plain python by building a dictionary of what index each value of the array would map to:
x = np.sort(np.unique(a))
index_dict = {j: i for i, j in enumerate(x)}
[index_dict[i] for i in a]
Seems you need to rank (dense) your array, in which case use scipy.stats.rankdata:
from scipy.stats import rankdata
rankdata(a, 'dense')-1
# array([ 0., 0., 1., 1., 2., 2., 2., 2., 3., 5., 4.])
This seems like it should be straightforward, but I can't figure it out.
Data source is a two column, comma delimited input file with these contents:
6,10
5,9
8,13
...
And my code is:
import numpy as np
data = np.loadtxt("data.txt", delimiter=",")
m = len(data)
x = np.reshape(data[:,0], (m,1))
y = np.ones((m,1))
z = np.matrix([x,y])
Which gives me this error:
Users/acpigeon/.virtualenvs/ipynb/lib/python2.7/site-packages/numpy-1.9.0.dev_297f54b-py2.7-macosx-10.9-intel.egg/numpy/matrixlib/defmatrix.pyc in __new__(subtype, data, dtype, copy)
270 shape = arr.shape
271 if (ndim > 2):
--> 272 raise ValueError("matrix must be 2-dimensional")
273 elif ndim == 0:
274 shape = (1, 1)
ValueError: matrix must be 2-dimensional
No amount of reshaping seems to get this to work, so I'm either missing something really simple or there's a better way to do this.
EDIT:
Would have been helpful to specify the output I am looking for. Here is a line of code that generates the desired result:
In [1]: np.matrix([[5,1],[6,1],[8,1]])
Out[1]:
matrix([[5, 1],
[6, 1],
[8, 1]])
The desired output can be generated this way:
In [12]: np.array((data[:, 0], np.ones(m))).transpose()
Out[12]:
array([[ 6., 1.],
[ 5., 1.],
[ 8., 1.]])
The above is copied from ipython and so has ipython style prompts.
Answer to previous version
To eliminate the error, replace:
x = np.reshape(data[:, 0], (m, 1))
with:
x = data[:, 0]
The former line produces a 2-dimensional matrix and that is what causes the error message. The latter produces a 1-D array with the same data.
Or how about first turning the array into a matrix, and then change the last column to 1?
In [2]: data=np.loadtxt('stack23859379.txt',delimiter=',')
In [3]: np.matrix(data)
Out[3]:
matrix([[ 6., 10.],
[ 5., 9.],
[ 8., 13.]])
In [4]: z = np.matrix(data)
In [5]: z[:,1]=1
In [6]: z
Out[6]:
matrix([[ 6., 1.],
[ 5., 1.],
[ 8., 1.]])