I have a 3D array (4,3,3) in which I would like to iteratively multiply with a 1D array (t variable) and sum to end up with an array (A) that is a summation of the four 3,3 arrays
I'm unsure on how I should be assigning indexes or how and if I should be using np.ndenumerate
Thanks
import numpy as np
import math
#Enter material constants for calculation of stiffness matrix
E1 = 20
E2 = 1.2
G12 = 0.8
v12=0.25
v21=(v12/E1)*E2
theta = np.array([30,-30,-30,30])
deg = ((math.pi*theta/180))
k = len(theta) #number of layers
t = np.array([0.005,0.005,0.005,0.005])
#Calculation of Q Values
Q11 = 1
Q12 = 2
Q21 = 3
Q22 = 4
Q66 = 5
Qbar = np.zeros((len(theta),3,3),order='F')
#CALCULATING THE VALUES OF THE QBAR MATRIX
for i, x in np.ndenumerate(deg):
m= np.cos(x) #sin of rotated lamina
n= np.sin(x) #cos of rotated lamina
Qbar11=Q11*3
Qbar12=Q22*4
Qbar16=Q16*4
Qbar21 = Qbar12
Qbar22=Q22*1
Qbar26=Q66*2
Qbar66=Q12*3
Qbar[i] = np.array([[Qbar11, Qbar12, Qbar16], [Qbar21, Qbar22, Qbar26], [Qbar16, Qbar26, Qbar66]], order = 'F')
print(Qbar)
A = np.zeros((3,3))
for i in np.nditer(t):
A[i]=Qbar[i]*t[i]
A=sum(A[i])
If I understand correctly, you want to multiply Qbar and t over the first axis, and then summing the result over the first axis (which results in an array of shape (3, 3)).
I created random arrays to make the code minimal:
import numpy as np
Qbar = np.random.randint(2, size=(4, 3, 3))
t = np.arange(4)
A = (Qbar * t[:, None, None]).sum(axis=0)
t[:, None, None] will create two new dimensions so that the shape becomes (4, 1, 1), which can be multiplied to Qbar element-wise. Then we just have to sum over the first axis.
NB: A = np.tensordot(t, Qbar, axes=([0],[0])) also works and can be faster for larger dimensions, but for the dimensions you provided I prefer the first solution.
Given that there is 2 Numpy Array :
3d_array with shape (100,10,2),
1d_indices with shape (100)
What is the Numpy way/equivalent to do this :
result = []
for i,j in zip(range(len(3d_array)),1d_indices):
result.append(3d_array[i,j])
Which should return result.shape (100,2)
The closest I've come to is by using fancy indexing on Numpy :
result = 3d_array[np.arange(len(3d_array)), 1d_indices]
Your code snippet should be equivalent to 3d_array[:, 1d_indices].reshape(-1,2), example:
a = np.arange(100*10*2).reshape(100,10,2) # 3d array
b = np.random.randint(0, 10, 100) # 1d indices
def fun(a,b):
result = []
for i in range(len(a)):
for j in b:
result.append(a[i,j])
return np.array(result)
assert (a[:, b].reshape(-1, 2) == fun(a, b)).all()
I'm trying to get the values of (nRows, nCols) from a 2D Matrix but when it's a single row (i.e. x = np.array([1, 2, 3, 4])), x.shape will return (4,) and so my statement of (nRows, nCols) = x.shape returns "ValueError: need more than 1 value to unpack"
Any suggestions on how I can make this statement more adaptable? It's for a function that is used in many programs and should work with both single row and multi-row matices. Thanks!
You could create a function that returns a tuple of rows and columns like this:
def rowsCols(a):
if len(a.shape) > 1:
rows = a.shape[0]
cols = a.shape[1]
else:
rows = a.shape[0]
cols = 0
return (rows, cols)
where a is the array you input to the function. Here's an example of using the function:
import numpy as np
x = np.array([1,2,3])
y = np.array([[1,2,3],[4,5,6],[7,8,9],[10,11,12]])
def rowsCols(a):
if len(a.shape) > 1:
rows = a.shape[0]
cols = a.shape[1]
else:
rows = a.shape[0]
cols = 0
return (rows, cols)
(nRows, nCols) = rowsCols(x)
print('rows {} and columns {}'.format(nRows, nCols))
(nRows, nCols) = rowsCols(y)
print('rows {} and columns {}'.format(nRows, nCols))
This prints rows 3 and columns 0 then rows 4 and columns 3. Alternatively, you can use the atleast_2d function for a more concise approach:
(r, c) = np.atleast_2d(x).shape
print('rows {}, cols {}'.format(r, c))
(r, c) = np.atleast_2d(y).shape
print('rows {}, cols {}'.format(r, c))
Which prints rows 1, cols 3 and rows 4, cols 3.
If your function uses
(nRows, nCols) = x.shape
it probably also indexes or iterates on x with the assumption that it has nRows rows, e.g.
x[0,:]
for row in x:
# do something with the row
Common practice is to reshape x (as needed) so it has at least 1 row. In other words, change the shape from (n,) to (1,n).
x = np.atleast_2d(x)
does this nicely. Inside a function, such a change to x won't affect x outside it. This way you can treat x as 2d through out your function, rather than constantly looking to see whether it is 1d v 2d.
Python: How can I force 1-element NumPy arrays to be two-dimensional?
is one of many previous SO questions that asks about treating 1d arrays as 2d.
I have a 2-D NumPy array and a set of indices the size of which is the first dimension of the NumPy array.
X = np.random.rand(5, 3)
a = np.random.randint(0, 3, 5)
I need to do something like
for i, ind in enumerate(a):
print X[i][ind]
Is there a vectorized way of doing this?
Here you go:
X = np.random.rand(5, 3)
a = np.random.randint(0, 3, 5)
In [12]: X[np.arange(a.size), a]
Out[12]: array([ 0.99653335, 0.30275346, 0.92844957, 0.54728781, 0.43535668])
In [13]: for i, ind in enumerate(a):
print X[i][ind]
# ....:
#0.996533345844
#0.30275345582
#0.92844956619
#0.54728781105
#0.435356681672
I'm assuming here that you don't need each value on a separate line and just want to extract the values.
I am trying to add one column to the array created from recfromcsv. In this case it's an array: [210,8] (rows, cols).
I want to add a ninth column. Empty or with zeroes doesn't matter.
from numpy import genfromtxt
from numpy import recfromcsv
import numpy as np
import time
if __name__ == '__main__':
print("testing")
my_data = recfromcsv('LIAB.ST.csv', delimiter='\t')
array_size = my_data.size
#my_data = np.append(my_data[:array_size],my_data[9:],0)
new_col = np.sum(x,1).reshape((x.shape[0],1))
np.append(x,new_col,1)
I think that your problem is that you are expecting np.append to add the column in-place, but what it does, because of how numpy data is stored, is create a copy of the joined arrays
Returns
-------
append : ndarray
A copy of `arr` with `values` appended to `axis`. Note that `append`
does not occur in-place: a new array is allocated and filled. If
`axis` is None, `out` is a flattened array.
so you need to save the output all_data = np.append(...):
my_data = np.random.random((210,8)) #recfromcsv('LIAB.ST.csv', delimiter='\t')
new_col = my_data.sum(1)[...,None] # None keeps (n, 1) shape
new_col.shape
#(210,1)
all_data = np.append(my_data, new_col, 1)
all_data.shape
#(210,9)
Alternative ways:
all_data = np.hstack((my_data, new_col))
#or
all_data = np.concatenate((my_data, new_col), 1)
I believe that the only difference between these three functions (as well as np.vstack) are their default behaviors for when axis is unspecified:
concatenate assumes axis = 0
hstack assumes axis = 1 unless inputs are 1d, then axis = 0
vstack assumes axis = 0 after adding an axis if inputs are 1d
append flattens array
Based on your comment, and looking more closely at your example code, I now believe that what you are probably looking to do is add a field to a record array. You imported both genfromtxt which returns a structured array and recfromcsv which returns the subtly different record array (recarray). You used the recfromcsv so right now my_data is actually a recarray, which means that most likely my_data.shape = (210,) since recarrays are 1d arrays of records, where each record is a tuple with the given dtype.
So you could try this:
import numpy as np
from numpy.lib.recfunctions import append_fields
x = np.random.random(10)
y = np.random.random(10)
z = np.random.random(10)
data = np.array( list(zip(x,y,z)), dtype=[('x',float),('y',float),('z',float)])
data = np.recarray(data.shape, data.dtype, buf=data)
data.shape
#(10,)
tot = data['x'] + data['y'] + data['z'] # sum(axis=1) won't work on recarray
tot.shape
#(10,)
all_data = append_fields(data, 'total', tot, usemask=False)
all_data
#array([(0.4374783740738456 , 0.04307289878861764, 0.021176067323686598, 0.5017273401861498),
# (0.07622262416466963, 0.3962146058689695 , 0.27912715826653534 , 0.7515643883001745),
# (0.30878532523061153, 0.8553768789387086 , 0.9577415585116588 , 2.121903762680979 ),
# (0.5288343561208022 , 0.17048864443625933, 0.07915689716226904 , 0.7784798977193306),
# (0.8804269791375121 , 0.45517504750917714, 0.1601389248542675 , 1.4957409515009568),
# (0.9556552723429782 , 0.8884504475901043 , 0.6412854758843308 , 2.4853911958174133),
# (0.0227638618687922 , 0.9295332854783015 , 0.3234597575660103 , 1.275756904913104 ),
# (0.684075052174589 , 0.6654774682866273 , 0.5246593820025259 , 1.8742119024637423),
# (0.9841793718333871 , 0.5813955915551511 , 0.39577520705133684 , 1.961350170439875 ),
# (0.9889343795296571 , 0.22830104497714432, 0.20011292764078448 , 1.4173483521475858)],
# dtype=[('x', '<f8'), ('y', '<f8'), ('z', '<f8'), ('total', '<f8')])
all_data.shape
#(10,)
all_data.dtype.names
#('x', 'y', 'z', 'total')
If you have an array, a of say 210 rows by 8 columns:
a = numpy.empty([210,8])
and want to add a ninth column of zeros you can do this:
b = numpy.append(a,numpy.zeros([len(a),1]),1)
The easiest solution is to use numpy.insert().
The Advantage of np.insert() over np.append is that you can insert the new columns into custom indices.
import numpy as np
X = np.arange(20).reshape(10,2)
X = np.insert(X, [0,2], np.random.rand(X.shape[0]*2).reshape(-1,2)*10, axis=1)
'''
np.append or np.hstack expects the appended column to be the proper shape, that is N x 1. We can use np.zeros to create this zeros column (or np.ones to create a ones column) and append it to our original matrix (2D array).
def append_zeros(x):
zeros = np.zeros((len(x), 1)) # zeros column as 2D array
return np.hstack((x, zeros)) # append column
I add a new column with ones to a matrix array in this way:
Z = append([[1 for _ in range(0,len(Z))]], Z.T,0).T
Maybe it is not that efficient?
It can be done like this:
import numpy as np
# create a random matrix:
A = np.random.normal(size=(5,2))
# add a column of zeros to it:
print(np.hstack((A,np.zeros((A.shape[0],1)))))
In general, if A is an m*n matrix, and you need to add a column, you have to create an n*1 matrix of zeros, then use "hstack" to add the matrix of zeros to the right of the matrix A.
Similar to some of the other answers suggesting using numpy.hstack, but more readable:
import numpy as np
# declare 10 rows x 3 cols integer array of all 1s
arr = np.ones((10, 3), dtype=np.int64)
# get the number of rows in the original array (as if we didn't know it was 10 or it could be different in other cases)
numRows = arr.shape[0]
# declare the new array which will be the new column, integer array of all 0s so it's visually distinct from the original array
additionalColumn = np.zeros((numRows, 1), dtype=np.int64)
# use hstack to tack on the additionl column
result = np.hstack((arr, additionalColumn))
print(result)
result:
$ python3 scratchpad.py
[[1 1 1 0]
[1 1 1 0]
[1 1 1 0]
[1 1 1 0]
[1 1 1 0]
[1 1 1 0]
[1 1 1 0]
[1 1 1 0]
[1 1 1 0]
[1 1 1 0]]
Here's a shorter one-liner:
import numpy as np
data = np.random.rand(210, 8)
data = np.c_[data, np.zeros(len(data))]
Something that I use often to convert points to homogenous coordinates with np.ones instead.