I began independently learning numpy using the numpy cookbook. I reviewed and executed the following code:
import scipy.misc
import matplotlib.pyplot
#This script demonstates fancy indexing by setting values
#On the diagnols to 0
#Load lena array
lena = scipy.misc.lena()
xmax = lena.shape[0]
ymax = lena.shape[1]
#Fancy indexing
#can set ranges of points to zero, all at once instead of using loop
lena[range(xmax), range(ymax)] = 0
lena[range(xmax-1,-1,-1), range(ymax)] = 0
matplotlib.pyplot.imshow(lena)
matplotlib.pyplot.show()
I understand everything in this code except:
lena[range(xmax), range(ymax)] = 0
lena[range(xmax-1,-1,-1), range(ymax)] = 0
I read the documentation on indexing and slicing but still cannot make sense of the above code. Here is are my points of confusion:
1)range(xmax) and range(ymax) encompass the entire x,y axes. Wouldn't setting them to zero make the entire image black?
2)What does range(xmax-1,-1,-1) mean?
Thanks guys!
The first bit of code is actually misleading, and relies on the fact that lena is a square image: what happens is equivalent to calling zip(range(xmax), range(ymax)), and then setting each of the resulting tuples to 0. You can see what could go wrong here: if xmax != ymax, then things won't work:
>>> test = lena[:,:-3]
>>> test.shape
(512, 509)
>>> xmax, ymax = test.shape
>>> test[range(xmax), range(ymax)] = 0
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ValueError: shape mismatch: objects cannot be broadcast to a single shape
It would probably be better to define diag_max = min(xmax, ymax), and then set lena[range(diag_max), range(diag_max)] = 0.
The answer to your second question is easier: range(from, to, step) is the general call to range:
>>> range(1, 10, 2)
[1, 3, 5, 7, 9]
>>> range(1, 10, -2)
[]
>>> range(10, 1, -2)
[10, 8, 6, 4, 2]
>>> range(10, 0, -1)
[10, 9, 8, 7, 6, 5, 4, 3, 2, 1]
In particular, this reverses the previous list, and so grabs the diagonal from right to left instead of left to right.
'range' will give you a list. Try it in your REPL and see what happens:
r = range(5)
# r is no [0,1,2,3,4]
So doing 'lena[range(xmax), range(ymax)] = 0' will set the diagonal of the 'lena' matrix to zero since you're stepping through the x and y coordinates incrementally at the same time.
'range' is quite simple. The answer from #JLLagrange answers it perfectly.
Related
I'm trying to make a Python app that shows a graph after the input of the data by the user, but the problem is that the y_array and the x_array do not have the same dimensions. When I run the program, this error is raised:
ValueError: x and y must have same first dimension, but have shapes () and ()
How can I draw a graph with the X and Y axis of different length?
Here is a minimal example code that will lead to the same error I got
:
import matplotlib.pyplot as plt
y = [0, 8, 9, 3, 0]
x = [1, 2, 3, 4, 5, 6, 7]
plt.plot(x, y)
plt.show()
This is virtually a copy/paste of the answer found here, but I'll show what I did to get these to match.
First, we need to decide which array to use- the x_array of length 7, or the y_array of length 5. I'll show both, starting with the former. Note that I am using numpy arrays, not lists.
Let's load the modules
import numpy as np
import matplotlib.pyplot as plt
import scipy.interpolate as interp
and the arrays
y = np.array([0, 8, 9, 3, 0])
x = np.array([1, 2, 3, 4, 5, 6, 7])
In both cases, we use interp.interp1d which is described in detail in the documentation.
For the x_array to be reduced to the length of the y_array:
x_inter = interp.interp1d(np.arange(x.size), x)
x_ = x_inter(np.linspace(0,x.size-1,y.size))
print(len(x_), len(y))
# Prints 5,5
plt.plot(x_,y)
plt.show()
Which gives
and for the y_array to be increased to the length of the x_array:
y_inter = interp.interp1d(np.arange(y.size), y)
y_ = y_inter(np.linspace(0,y.size-1,x.size))
print(len(x), len(y_))
# Prints 7,7
plt.plot(x,y_)
plt.show()
Which gives
I'll try to explain my issue here without going into too much detail on the actual application so that we can stay grounded in the code. Basically, I need to do operations to a vector field. My first step is to generate the field as
x,y,z = np.meshgrid(np.linspace(-5,5,10),np.linspace(-5,5,10),np.linspace(-5,5,10))
Keep in mind that this is a generalized case, in the program, the bounds of the vector field are not all the same. In the general run of things, I would expect to say something along the lines of
u,v,w = f(x,y,z).
Unfortunately, this case requires so more difficult operations. I need to use a formula similar to
where the vector r is defined in the program as np.array([xgrid-x,ygrid-y,zgrid-z]) divided by its own norm. Basically, this is a vector pointing from every point in space to the position (x,y,z)
Now Numpy has implemented a cross product function using np.cross(), but I can't seem to create a "meshgrid of vectors" like I need.
I have a lambda function that is essentially
xgrid,ygrid,zgrid=np.meshgrid(np.linspace(-5,5,10),np.linspace(-5,5,10),np.linspace(-5,5,10))
B(x,y,z) = lambda x,y,z: np.cross(v,np.array([xgrid-x,ygrid-y,zgrid-z]))
Now the array v is imported from another class and seems to work just fine, but the second array, np.array([xgrid-x,ygrid-y,zgrid-z]) is not a proper shape because it is a "vector of meshgrids" instead of a "meshgrid of vectors". My big issue is that I cannot seem to find a method by which to format the meshgrid in such a way that the np.cross() function can use the position vector. Is there a way to do this?
Originally I thought that I could do something along the lines of:
x,y,z = np.meshgrid(np.linspace(-2,2,5),np.linspace(-2,2,5),np.linspace(-2,2,5))
A = np.array([x,y,z])
cross_result = np.cross(np.array(v),A)
This, however, returns the following error, which I cannot seem to circumvent:
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "C:\Python27\lib\site-packages\numpy\core\numeric.py", line 1682, in cross
raise ValueError(msg)
ValueError: incompatible dimensions for cross product
(dimension must be 2 or 3)
There's a work around with reshape and broadcasting:
A = np.array([x_grid, y_grid, z_grid])
# A.shape == (3,5,5,5)
def B(v, p):
'''
v.shape = (3,)
p.shape = (3,)
'''
shape = A.shape
Ap = A.reshape(3,-1) - p[:,None]
return np.cross(v[None,:], Ap.reshape(3,-1).T).reshape(shape)
print(B(v,p).shape)
# (3, 5, 5, 5)
I think your original attempt only lacks the specification of the axis along which the cross product should be executed.
x, y, z = np.meshgrid(np.linspace(-2, 2, 5),np.linspace(-2, 2, 5), np.linspace(-2, 2, 5))
A = np.array([x, y, z])
cross_result = np.cross(np.array(v), A, axis=0)
I tested this with the code below. As an alternative to np.array([x, y, z]), you can also use np.stack(x, y, z, axis=0), which clearly shows along which axis the meshgrids are stacked to form a meshgrid of vectors, the vectors being aligned with axis 0. I also printed the shape each time and used random input for testing. In the test, the output of the formula is compared at a random index to the cross product of the input-vector at the same index with vector v.
import numpy as np
x, y, z = np.meshgrid(np.linspace(-5, 5, 10), np.linspace(-5, 5, 10), np.linspace(-5, 5, 10))
p = np.random.rand(3) # random reference point
A = np.array([x-p[0], y-p[1], z-p[2]]) # vectors from positions to reference
A_bis = np.stack((x-p[0], y-p[1], z-p[2]), axis=0)
print(f"A equals A_bis? {np.allclose(A, A_bis)}") # the two methods of stacking yield the same
v = -1 + 2*np.random.rand(3) # random vector v
B = np.cross(v, A, axis=0) # cross-product for all points along correct axis
print(f"Shape of v: {v.shape}")
print(f"Shape of A: {A.shape}")
print(f"Shape of B: {B.shape}")
print("\nComparison for random locations: ")
point = np.random.randint(0, 9, 3) # generate random multi-index
a = A[:, point[0], point[1], point[2]] # look up input-vector corresponding to index
b = B[:, point[0], point[1], point[2]] # look up output-vector corresponding to index
print(f"A[:, {point[0]}, {point[1]}, {point[2]}] = {a}")
print(f"v = {v}")
print(f"Cross-product as v x a: {np.cross(v, a)}")
print(f"Cross-product from B (= v x A): {b}")
The resulting output looks like:
A equals A_bis? True
Shape of v: (3,)
Shape of A: (3, 10, 10, 10)
Shape of B: (3, 10, 10, 10)
Comparison for random locations:
A[:, 8, 1, 1] = [-4.03607312 3.72661831 -4.87453077]
v = [-0.90817859 0.10110274 -0.17848181]
Cross-product as v x a: [ 0.17230515 -3.70657882 -2.97637688]
Cross-product from B (= v x A): [ 0.17230515 -3.70657882 -2.97637688]
I newbie in python. I have matlab script like below. I want to re-write matrix 3D in matlab script in to python 3.x language. How can I fix it?
nl=length(res);
ndat=length(per);
phi=atan(1)*4;
amu=phi*4e-7;
for i=1:ndat
for j=1:nl
z=sqrt(phi*amu*res(j)/per(i));
zz(j)=complex(z,z);
exp0=exp((-2)*zz(j)/res(j)*thi(j));
exp1=complex(1,0)+exp0;
exp2=complex(1,0)-exp0;
%matrix 3D
ldi(1,1,j)=exp1;
ldi(1,2,j)=zz(j)*exp2
ldi(2,1,j)=exp2/zz(j);
ldi(2,2,j)=exp1;`
end
end
You'll find a self-contained implementation of your code (below), with a few key differences:
Python indexing starts from 0 instead of one
Python indexing uses square brackets instead of round ones
Mathematic functions must be imported from libraries (here math and cmath)
Good luck!
import math
import cmath
# Data
res = [1, 4, 1, 2, 3]
per = [5, 5, 1, 1, 0.5, 0.6]
thi = [1, 2, 3, 4, 5, 6]
nl = len(res)
ldi = [[[0 for x in range(nl)],[0 for x in range(nl)]], [[0 for x in range(nl)],[0 for x in range(nl)]]]
zz = [0]*nl
nl = len(res)
ndat = len(per)
phi = math.atan(1)*4
amu = phi*4e-7
for i in range(ndat):
for j in range(nl):
z = math.sqrt(phi*amu*res[j]/per[i])
zz[j] = complex(z,z)
exp0=cmath.exp((-2)*zz[j]/res[j]*thi[j]);
exp1=complex(1,0)+exp0;
exp2=complex(1,0)-exp0;
#- matrix 3D
ldi[0][0][j]=exp1;
ldi[0][1][j]=zz[j]*exp2
ldi[1][0][j]=exp2/zz[j]
ldi[1][1][j]=exp1
Something very simple in Matlab, but I can't get it in Python. How to get the following:
x=np.array([1,2,3])
y=np.array([4,5,6,7])
z=x.T*y
z=
[[4,5,6,7],
[8,10,12,14],
[12,15,18,21]]
As in
x [4][5][6][7]
[1]
[2]
[3]
In scientific python that would be an outer product np.outer(x,y)
See http://docs.scipy.org/doc/numpy/reference/generated/numpy.outer.html:
import numpy;
>>> x=numpy.array([1,2,3])
>>> y=numpy.array([4,5,6,7])
>>> numpy.outer(x,y)
array([[ 4, 5, 6, 7],
[ 8, 10, 12, 14],
[12, 15, 18, 21]])
In MATLAB, size(x) is (1,3). So x' is (3,1). Multiply that by y, which is (1,4), produces (3,4) shape.
In numpy, x.shape is (3,). x.T is the same. So to get the same outer product, you need to expand the dimensions of x and y. One way is with reshape.
z = x.reshape(3,1)* y.reshape(1,4)
numpy also lets you do this with a newaxis indexing (None also works). It also automatically adds beginning newaxis if that is needed. So this also does the job:
z = x[:,np.newaxis]*y
np.outer does exactly this (with a minor embelishment): a.ravel()[:, newaxis]*b.ravel()[newaxis,:].
There's another tool in numpy
z = np.einsum('i,j->ij',x,y)
It is based on an indexing notation that is popular in physics, and is especially useful in writing more complicated inner (dot) products.
Using list comprehension:
x = [1, 2, 3]
y = [4, 5, 6, 7]
z = [[i * j for j in y] for i in x]
I'm trying to slice and iterate over a multidimensional array at the same time. I have a solution that's functional, but it's kind of ugly, and I bet there's a slick way to do the iteration and slicing that I don't know about. Here's the code:
import numpy as np
x = np.arange(64).reshape(4,4,4)
y = [x[i:i+2,j:j+2,k:k+2] for i in range(0,4,2)
for j in range(0,4,2)
for k in range(0,4,2)]
y = np.array(y)
z = np.array([np.min(u) for u in y]).reshape(y.shape[1:])
Your last reshape doesn't work, because y has no shape defined. Without it you get:
>>> x = np.arange(64).reshape(4,4,4)
>>> y = [x[i:i+2,j:j+2,k:k+2] for i in range(0,4,2)
... for j in range(0,4,2)
... for k in range(0,4,2)]
>>> z = np.array([np.min(u) for u in y])
>>> z
array([ 0, 2, 8, 10, 32, 34, 40, 42])
But despite that, what you probably want is reshaping your array to 6 dimensions, which gets you the same result as above:
>>> xx = x.reshape(2, 2, 2, 2, 2, 2)
>>> zz = xx.min(axis=-1).min(axis=-2).min(axis=-3)
>>> zz
array([[[ 0, 2],
[ 8, 10]],
[[32, 34],
[40, 42]]])
>>> zz.ravel()
array([ 0, 2, 8, 10, 32, 34, 40, 42])
It's hard to tell exactly what you want in the last mean, but you can use stride_tricks to get a "slicker" way. It's rather tricky.
import numpy.lib.stride_tricks
# This returns a view with custom strides, x2[i,j,k] matches y[4*i+2*j+k]
x2 = numpy.lib.stride_tricks(
x, shape=(2,2,2,2,2,2),
strides=(numpy.array([32,8,2,16,4,1])*x.dtype.itemsize))
z2 = z2.min(axis=-1).min(axis=-2).min(axis=-3)
Still, I can't say this is much more readable. (Or efficient, as each min call will make temporaries.)
Note, my answer differs from Jaime's because I tried to match your elements of y. You can tell if you replace the min with max.