If I reshape in python I use this:
import numpy as np
y= np.asarray([1,2,3,4,5,6,7,8])
x=2
z=y.reshape(-1, x)
print(z)
and get this
>>>
[[1 2]
[3 4]
[5 6]
[7 8]]
How would I get the same thing in julia? I tried:
z = [1,2,3,4,5,6,7,8]
x= 2
a=reshape(z,x,4)
println(a)
and it gave me:
[1 3 5 7
2 4 6 8]
If I use reshape(z,4,x) it would give
[1 5
2 6
3 7
4 8]
Also is there a way to do reshape without specifying the second dimension like reshape(z,x) or if the secondary dimension is more ambiguous?
I think what you have hit upon is NumPy stores in row-major order and Julia stores arrays in column major order as covered here.
So Julia is doing what numpy would do if you used
z=y.reshape(-1,x,order='F')
what you want is the transpose of your first attempt, which is
z = [1,2,3,4,5,6,7,8]
x= 2
a=reshape(z,x,4)'
println(a)
you want to know if there is something that will compute the 2nd dimension assuming the array is 2 dimensional? Not that I know of. Possibly ArrayViews? Here's a simple function to start
julia> shape2d(x,shape...)=length(shape)!=1?reshape(x,shape...):reshape(x,shape[1],Int64(length(x)/shape[1]))
shape2d (generic function with 1 method)
julia> shape2d(z,x)'
4x2 Array{Int64,2}:
1 2
3 4
5 6
7 8
How about
z = [1,2,3,4,5,6,7,8]
x = 2
a = reshape(z,x,4)'
which gives
julia> a = reshape(z,x,4)'
4x2 Array{Int64,2}:
1 2
3 4
5 6
7 8
As for your bonus question
"Also is there a way to do reshape without specifying the second
dimension like reshape(z,x) or if the secondary dimension is more
ambiguous?"
the answer is not exactly, because it'd be ambiguous: reshape can make 3D, 4D, ..., tensors so its not clear what is expected. You can, however, do something like
matrix_reshape(z,x) = reshape(z, x, div(length(z),x))
which does what I think you expect.
"Also is there a way to do reshape without specifying the second dimension like reshape(z,x) or if the secondary dimension is more ambiguous?"
Use : instead of -1
I'm using Julia 1.1 (not sure if there was a feature when it was originally answered)
julia> z = [1,2,3,4,5,6,7,8]; a = reshape(z,:,2)
4×2 Array{Int64,2}:
1 5
2 6
3 7
4 8
However, if you want the first row to be 1 2 and match Python, you'll need to follow the other answer mentioning row-major vs column-major ordering and do
julia> z = [1,2,3,4,5,6,7,8]; a = reshape(z,2,:)'
4×2 LinearAlgebra.Adjoint{Int64,Array{Int64,2}}:
1 2
3 4
5 6
7 8
Related
this is my list :
N= 9
Mylist=[9,8,7,6,5,4,3,2,1]
For this input
Output should be :
9 8 7
6 5 4
3 2 1
It sounds like you're wondering how to turn a list into a numpy array of a particular shape. Documentation is here.
import numpy as np
my_list=[3,9,8,7,6,5,4,3,2,1]
# Dropping the first item of your list as it isn't used in your output
array = np.array(my_list[1:]).reshape((3,3))
print(array)
Output
[[9 8 7]
[6 5 4]
[3 2 1]]
Is there a way to build the Dot product of two matrices with different shape, without using anything else as pure python and numpy?
The shape of the columns should be equal, but the rows should be different. (example below)
Of course I know the brute force way:
for i in A:
for j in B:
np.dot(A,B)
but is there something else?
Here an example:
import numpy as np
A = np.full((4,5),3)
B = np.full((3,5),5)
print(A)
print(B)
result = np.zeros((A.shape[0],B.shape[0]))
for i in range(A.shape[0]):
for j in range(B.shape[0]):
result[i,j] = np.dot(A[i],B[j])
print(dot)
Output:
A = [[3 3 3 3 3]
[3 3 3 3 3]
[3 3 3 3 3]
[3 3 3 3 3]]
B = [[5 5 5 5 5]
[5 5 5 5 5]
[5 5 5 5 5]]
result = [[75. 75. 75.]
[75. 75. 75.]
[75. 75. 75.]
[75. 75. 75.]]
The coal is to calculate the dot product without two loops. So is there a more efficient way?
I have a list of four strings. Then in a Pandas dataframe I want to create a variable randomly selecting a value from this list and assign into each row. I am using numpy's random choice, but reading their documentation, there is no seed option. How can I specify the random seed to the random assignment so every time the random assignment will be the same?
service_code_options = ['899.59O', '12.42R', '13.59P', '204.68L']
df['SERVICE_CODE'] = [np.random.choice(service_code_options ) for i in df.index]
You need define it before by numpy.random.seed, also list comprehension is not necessary, because is possible use numpy.random.choice with parameter size:
np.random.seed(123)
df = pd.DataFrame({'a':range(10)})
service_code_options = ['899.59O', '12.42R', '13.59P', '204.68L']
df['SERVICE_CODE'] = np.random.choice(service_code_options, size=len(df))
print (df)
a SERVICE_CODE
0 0 13.59P
1 1 12.42R
2 2 13.59P
3 3 13.59P
4 4 899.59O
5 5 13.59P
6 6 13.59P
7 7 12.42R
8 8 204.68L
9 9 13.59P
Documentation numpy.random.seed
np.random.seed(this_is_my_seed)
That could be an integer or a list of integers
np.random.seed(300)
Or
np.random.seed([3, 1415])
Example
np.random.seed([3, 1415])
service_code_options = ['899.59O', '12.42R', '13.59P', '204.68L']
np.random.choice(service_code_options, 3)
array(['899.59O', '204.68L', '13.59P'], dtype='<U7')
Notice that I passed a 3 to the choice function to specify the size of the array.
numpy.random.choice
According to the notes of numpy.random.seed in numpy v1.2.4:
Best practice is to use a dedicated Generator instance rather than the random variate generation methods exposed directly in the random module.
Such a Generator is constructed using np.random.default_rng.
Thus, instead of np.random.seed, the current best practice is to use a np.random.default_rng with a seed to construct a Generator, which can be further used for reproducible results.
Combining jezrael's answer and the current best practice, we have:
import pandas as pd
import numpy as np
rng = np.random.default_rng(seed=121)
df = pd.DataFrame({'a':range(10)})
service_code_options = ['899.59O', '12.42R', '13.59P', '204.68L']
df['SERVICE_CODE'] = rng.choice(service_code_options, size=len(df))
print(df)
a SERVICE_CODE
0 0 12.42R
1 1 13.59P
2 2 12.42R
3 3 12.42R
4 4 899.59O
5 5 204.68L
6 6 204.68L
7 7 13.59P
8 8 12.42R
9 9 13.59P
Consider that I have some vectors (or strings) of numbers which generally have different length, eg x=[1 2 3 4 3 3 3 2 5].
Now, for a new vector y I want to find which one of the existing vectors x is the most similar.
Any idea?
The complete Problem:
I want to predict a time serie with some Neural Networks. Atevery step all the networks predict the next value of the serie. When the real value comes, the network that did the best prediction wins and I write its number to the vector X. After i finish with the timeserie1 i will have generate a vector X1 and each element of it will represend the best NN.
Now consider that i have 10 time series , so 10 X vectors. For a new one time serie Y i will do the same procedure. I want to define the kind of the Y using its similarity between this and the X vectors. I think the most importand aspect is the succession of the NN. I need for output something like a number or percentage of similarity.
eg:
X1= [ 1 1 2 2 3 3 4 4 5 5 6 6 ]
X2=[1 2 3 4 5 6 1 2 3 4 5 6]
Y=[1 1 1 2 2 3 4 5 5 6 6]
Then Y is more similar to X1
I have n sets, each having n1,n2,n3...nN distinct members.
How do I generate n1*n2*n3...*nN possible combinations from them
e.g
[6] [4 5] [1 2 3 4]
will give
6 4 1
6 4 2
6 4 3
6 4 4
6 5 1
6 5 2
6 5 3
6 5 4
I want to do this in matlab, but a normal algorithm would also be fine
An easy solution is to simulate a sum !
Start with a list of indices 0 0 0, corresponding to the indices of your values. That leads you to the value 6 4 1 in your example.
then add 1.
You now have indices 001, so 642
and so on.
at 004, you overflow, so your indices become 010, having 6 5 1
Keep doing that, and keep a counter of the visited possibilites. There are 1 * 2 * 4 possibilities, so it's easy to know when you are done.
I think you're looking for Cartesian product of sets:
This should help:
cartprod(N1,N2,N3, ...)
http://www.mathworks.com/matlabcentral/fileexchange/5475-cartprod-cartesian-product-of-multiple-sets
There's another one here
set = {n1, n2, n3, ...}
allcomb(set{:})