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Assinging submatrix to matrix in numpy [duplicate]

This question already has an answer here:
Numpy getting in the way of int -> float type casting
(1 answer)
Closed 2 years ago.
I wrote a some code to compute householder reduction to Hessenberg form
V = []
m,n = A.shape
for i in range(m-1):
x = A[i+1:,i]
e1 = np.zeros(x.shape)
e1[0] = 1
v = sgn(x[0])*np.linalg.norm(x)*e1 + x
v = v/np.linalg.norm(v)
V.append(v)
vv = np.outer(v,v)
print(A[i+1:,i:]-2*vv # A[i+1:,i:])
A[i+1:,i:] =A[i+1:,i:]-2*vv # A[i+1:,i:]
print(A)
A[:,i+1:] = A[:,i+1:] - 2 * np.outer(A[:,i+1:] # v,v)
I run this code with A =
[[1,2,3],
[2,4,5],
[1,3,2]]
The first print statement prints
[[-2.23606798 -4.91934955 -5.36656315]
[ 0. 0.89442719 -0.4472136 ]]
Which is what makes sense.
While the second prints
[[ 1 2 3]
[-2 -4 -5]
[ 0 0 0]]
And this don't make sense.
Why do they print differently?
If this kind of assignment don't work, is there some other smart way?

You are trying to assign floats to an array of ints. Here is a quick fix - Change the dtype before operating on it.
A = np.array([[1,2,3],
[2,4,5],
[1,3,2]])
b = np.array([[-2.23606798, -4.91934955, -5.36656315],
[ 0., 0.89442719, -0.4472136 ]])
A = A.astype('float64')
A[1:,0:] = b
print(A)
>>>
[[ 1. 2. 3. ]
[-2.23606798 -4.91934955 -5.36656315]
[ 0. 0.89442719 -0.4472136 ]]
>>>
From the documentation regarding indexing/assignments:
Note that assignments may result in changes if assigning higher types to lower types (like floats to ints) or even exceptions (assigning complex to floats or ints):

Checking for a Magic Square Python

i'm trying to make a function that checks whether or not a matrix is a magic square. i only need to check the vertical and horizontal (not diagonal). sometimes it passes and sometimes it fails. i was hoping that someone could help me solve the issue. this is my code:
for i in range(len(m)):
if len(m[i]) != len(m):
return False
return True
and this is the one that fails. it returns false:
m = [ [1,2,3,4]
, [5,6,7,8]
, [9,10,11,12]
, [13,14,15,16]
]
print(magic_square(m)) == False)

The code you provide does not check if a matrix is a magic square, it only checks if a matrix (in this case, a list of lists) is square. After this check, you need to calculate the sums of each row, each column and each diagonal (although for some reason you said you don't need those) and compare if all of them are equal.

If you are ok with using numpy then you can use
m = np.array(m)
len(np.unique(np.concatenate([m.sum(axis=1), m.sum(axis=0)]))) == 1
Test cases:
m = [ [1,2,3,4]
, [5,6,7,8]
, [9,10,11,12]
, [13,14,15,16]
]
m = np.array(m)
print (len(np.unique(np.concatenate([m.sum(axis=1), m.sum(axis=0)]))) == 1)
m = [ [2,7,6]
, [9,5,1]
, [4,3,8]
]
m = np.array(m)
print (len(np.unique(np.concatenate([m.sum(axis=1), m.sum(axis=0)]))) == 1)
Output:
False
True
Meaning:
m.sum(axis=1) : Sum the numpy array along the rows
m.sum(axis=0) : Sum the numpy array along the columns
np.concatenate([m.sum(axis=1), m.sum(axis=0)]) : Combine both the sums (along rows and columns) into t single numpy array
np.unique(x) : Find the number of unique elements in the numpy array x
len(np.unique(np.concatenate([m.sum(axis=1), m.sum(axis=0)]))) == 1) : Check the number of unique elements in the row wise sum and columns wise sum is 1. i.e all the row wise sum and column wise sums are same.

This is not as clever as the numpy answer, but it works
def magic_square(m):
# check size
for i in range(len(m)):
if len(m[i]) != len(m):
return False
# check row sums
for r in m:
if sum(r) != sum(m[0]):
return False
# check column sums
cols = [[r[c] for r in m] for c in range(len(m[0]))]
for c in cols:
if sum(c) != sum(m[0]):
return False
return True
m = [ [1,2,3,4]
, [5,6,7,8]
, [9,10,11,12]
, [13,14,15,16]
]
print(magic_square(m)) # False
m = [ [8,11,14,1]
, [13,2,7,12]
, [3,16,9,6]
, [10,5,4,15]
]
print(magic_square(m)) # True

How to tabulate several vertical arrays in python?

I have used the following "for loop" to generate a series of column vectors
for x in time:
S = dot(M, S)
print S
Where M is (n x n) matrix and S is (n x 1) matrix. I am therefore trying to find the product of these two matrices.
The result is displayed as:
I would like the result to be displayed in a table like this.
The number of column vectors is not limited to 4, rather it is "n".

If you really need each iteration as a new column, I guess you will need to accumulate the results in a 2d array and print at the end. For example:
import numpy as np
M = np.random.randn(5,5)
S = np.random.randn(5,1)
for x in range(4):
S = np.c_[S, np.dot(M, S[:,-1])]
np.set_printoptions(precision=5, linewidth=120)
print S
which prints:
[[ 0.19891 -0.46714 2.09736 -5.01507 14.7212 ]
[ 0.6387 0.81975 -2.25251 6.5098 -8.27462]
[ -0.44047 0.3941 1.81101 -7.24052 23.07632]
[ -0.17742 0.88452 -2.80172 10.06426 -21.50157]
[ 0.57601 -0.80838 1.14127 1.15622 -4.11907]]

Python Numpy - Square Values Issue

I'm trying to square all the elements in a numpy array but the results are not what I'm expecting (ie some are negative numbers and none are the actual square values). Can anyone please explain what I'm doing wrong and/or whats going on?
import numpy as np
import math
f = 'file.bin'
frameNum = 25600
channelNum = 2640
data = np.fromfile(f,dtype=np.int16)
total = frameNum*channelNum*2
rs = data[:total].reshape(channelNum,-1) #reshaping the data a little. Omitting added values at the end.
I = rs[:,::2] # pull out every other column
print "Shape :", I.shape
print "I : ", I[1,:10]
print "I**2 : ", I[1,:10]**2
print "I*I : ",I[1,:10]* I[1,:10]
print "np.square : ",np.square(I[1,:10])
exit()
Output:
Shape : (2640L, 25600L)
I : [-5302 -5500 -5873 -5398 -5536 -6708 -6860 -6506 -6065 -6363]
I**2 : [ -3740 -27632 20193 -25116 -23552 -25968 4752 -8220 18529 -13479]
I*I : [ -3740 -27632 20193 -25116 -23552 -25968 4752 -8220 18529 -13479]
np.square : [ -3740 -27632 20193 -25116 -23552 -25968 4752 -8220 18529 -13479]
Any suggestions?

It is because of the dtype=np.int16. You are allowing only 16 bits to represent the numbers, and -5302**2 is larger than the maximum value (32767) that a signed 16-bit integer can take. So you're seeing only the lowest 16 bits of the result, the first of which is interpreted (or, from your point of view, misinterpreted) as a sign bit.
Convert your array to a different dtype - for example
I = np.array( I, dtype=np.int32 )
or
I = np.array( I, dtype=np.float )
before performing numerical operations that might go out of range.
With dtype=np.int16, the highest-magnitude integers you can square are +181 and -181. The square of 182 is larger than 32767 and so it overflows. Even with dtype=np.int32 representation, the highest-magnitude integers you can square are +46340 and -46340: the square of 46341 overflows.

This is the reason:
>>> a = np.array([-5302, -5500], dtype=np.int16)
>>> a * a
array([ -3740, -27632], dtype=int16)
This the solution:
b = np.array([-5302, -5500], dtype=np.int32)
>>> b * b
>>> array([28111204, 30250000], dtype=int32)
Change:
data = np.fromfile(f, dtype=np.int16)
into:
data = np.fromfile(f, dtype=np.in16).astype(np.int32)

Applying several functions to each row of an array

I have a numpy array which has only a few non-zero entries which can be either positive or negative. E.g. something like this:
myArray = np.array([[ 0. , 0. , 0. ],
[ 0.32, -6.79, 0. ],
[ 0. , 0. , 0. ],
[ 0. , 1.5 , 0. ],
[ 0. , 0. , -1.71]])
In the end, I would like to receive a list where each entry of this list corresponds to a row of myArray and is a cumulative product of function outputs which depend on the entries of the respective row of myArray and another list (in the example below it is called l).
The individual terms depend on the sign of the myArray entry: When it is positive, I apply "funPos", when it is negative, I apply "funNeg" and if the entry is 0, the term will be 1. So in the example array from above it would be:
output = [1*1*1 ,
funPos(0.32, l[0])*funNeg(-6.79,l[1])*1,
1*1*1,
1*funPos(1.5, l[1])*1,
1*1*funNeg(-1.71, l[2])]
I implemented this as shown below and it gives me the desired output (note: that is just a highly simplified toy example; the actual matrices are far bigger and the functions more complicated). I go through each row of the array, if the sum of the row is 0, I don't have to do any calculations and the output is just 1. If it is not equal 0, I go through this row, check the sign of each value and apply the appropriate function.
import numpy as np
def doCalcOnArray(Array1, myList):
output = np.ones(Array1.shape[0]) #initialize output
for indRow,row in enumerate(Array1):
if sum(row) != 0: #only then calculations are needed
tempProd = 1. #initialize the product that corresponds to the row
for indCol, valCol in enumerate(row):
if valCol > 0:
tempVal = funPos(valCol, myList[indCol])
elif valCol < 0:
tempVal = funNeg(valCol, myList[indCol])
elif valCol == 0:
tempVal = 1
tempProd = tempProd*tempVal
output[indRow] = tempProd
return output
def funPos(val1,val2):
return val1*val2
def funNeg(val1,val2):
return val1*(val2+1)
myArray = np.array([[ 0. , 0. , 0. ],
[ 0.32, -6.79, 0. ],
[ 0. , 0. , 0. ],
[ 0. , 1.5 , 0. ],
[ 0. , 0. , -1.71]])
l = [1.1, 2., 3.4]
op = doCalcOnArray(myArray,l)
print op
The output is
[ 1. -7.17024 1. 3. -7.524 ]
which is the desired one.
My question is whether there is a more efficient way for doing that since that is quite "expensive" for large arrays.
EDIT:
I accepted gabhijit's answer because the pure numpy solution he came up with seems to be the fastest one for the arrays I am dealing with. Please note, that there is also a nice working solution from RaJa that requires panda and also the solution from dave works fine which can serve as a nice example on how to use generators and numpy's "apply_along_axis".

Here's what I have tried - using reduce, map. I am not sure how fast this is - but is this what you are trying to do?
Edit 4: Simplest and most readable - Make l a numpy array and then greatly simplifies where.
import numpy as np
import time
l = np.array([1.0, 2.0, 3.0])
def posFunc(x,y):
return x*y
def negFunc(x,y):
return x*(y+1)
def myFunc(x, y):
if x > 0:
return posFunc(x, y)
if x < 0:
return negFunc(x, y)
else:
return 1.0
myArray = np.array([
[ 0.,0.,0.],
[ 0.32, -6.79, 0.],
[ 0.,0.,0.],
[ 0.,1.5,0.],
[ 0.,0., -1.71]])
t1 = time.time()
a = np.array([reduce(lambda x, (y,z): x*myFunc(z,l[y]), enumerate(x), 1) for x in myArray])
t2 = time.time()
print (t2-t1)*1000000
print a
Basically let's just look at last line it says cumulatively multiply things in enumerate(xx), starting with 1 (last parameter to reduce). myFunc simply takes the element in myArray(row) and element # index row in l and multiplies them as needed.
My output is not same as yours - so I am not sure whether this is exactly what you want, but may be you can follow the logic.
Also I am not so sure how fast this will be for huge arrays.
edit: Following is a 'pure numpy way' to do this.
my = myArray # just for brevity
t1 = time.time()
# First set the positive and negative values
# complicated - [my.itemset((x,y), posFunc(my.item(x,y), l[y])) for (x,y) in zip(*np.where(my > 0))]
# changed to
my = np.where(my > 0, my*l, my)
# complicated - [my.itemset((x,y), negFunc(my.item(x,y), l[y])) for (x,y) in zip(*np.where(my < 0))]
# changed to
my = np.where(my < 0, my*(l+1), my)
# print my - commented out to time it.
# Now set the zeroes to 1.0s
my = np.where(my == 0.0, 1.0, my)
# print my - commented out to time it
a = np.prod(my, axis=1)
t2 = time.time()
print (t2-t1)*1000000
print a
Let me try to explain the zip(*np.where(my != 0)) part as best as I can. np.where simply returns two numpy arrays first array is an index of row, second array is an index of column that matches the condition (my != 0) in this case. We take a tuple of those indices and then use array.itemset and array.item, thankfully, column index is available for free to us, so we can just take the element # that index in the list l. This should be faster than previous (and by orders of magnitude readable!!). Need to timeit to find out whether it indeed is.
Edit 2: Don't have to call separately for positive and negative can be done with one call np.where(my != 0).

So, let's see if I understand your question.
You want to map elements of your matrix to a new matrix such that:
0 maps to 1
x>0 maps to funPos(x)
x<0 maps to funNeg(x)
You want to calculate the product of all elements in the rows this new matrix.
So, here's how I would go about doing it:
1:
def myFun(a):
if a==0:
return 1
if a>0:
return funPos(a)
if a<0:
return funNeg(a)
newFun = np.vectorize(myFun)
newArray = newFun(myArray)
And for 2:
np.prod(newArray, axis = 1)
Edit: To pass the index to funPos, funNeg, you can probably do something like this:
# Python 2.7
r,c = myArray.shape
ctr = -1 # I don't understand why this should be -1 instead of 0
def myFun(a):
global ctr
global c
ind = ctr % c
ctr += 1
if a==0:
return 1
if a>0:
return funPos(a,l[ind])
if a<0:
return funNeg(a,l[ind])

I think this numpy function would be helpful to you
numpy.apply_along_axis
Here is one implementation. Also I would warn against checking if the sum of the array is 0. Comparing floats to 0 can give unexpected behavior due to machine accuracy constraints. Also if you have -5 and 5 the sum is zero and I'm not sure thats what you want. I used numpy's any() function to see if anything was nonzero. For simplicity I also pulled your list (my_list) into global scope.
import numpy as np
my_list = 1.1, 2., 3.4
def func_pos(val1, val2):
return val1 * val2
def func_neg(val1, val2):
return val1 *(val2 + 1)
def my_generator(row):
for i, a in enumerate(row):
if a > 0:
yield func_pos(a, my_list[i])
elif a < 0:
yield func_neg(a, my_list[i])
else:
yield 1
def reduce_row(row):
if not row.any():
return 1.0
else:
return np.prod(np.fromiter(my_generator(row), dtype=float))
def main():
myArray = np.array([
[ 0. , 0. , 0. ],
[ 0.32, -6.79, 0. ],
[ 0. , 0. , 0. ],
[ 0. , 1.5 , 0. ],
[ 0. , 0. , -1.71]])
return np.apply_along_axis(reduce_row, axis=1, arr=myArray)
There are probably faster implmentations, I think apply_along_axis is really just a loop under the covers.
I didn't test, but I bet this is faster than what you started with, and should be more memory efficient.

I've tried your example with the masking function of numpy arrays. However, I couldn't find a solution to replace the values in your array by funPos or funNeg.
So my suggestion would be to try this using pandas instead as it conserves indices while masking.
See my example:
import numpy as np
import pandas as pd
def funPos(a, b):
return a * b
def funNeg(a, b):
return a * (b + 1)
myPosFunc = np.vectorize(funPos) #vectorized form of funPos
myNegFunc = np.vectorize(funNeg) #vectorized form of funNeg
#Input
I = [1.0, 2.0, 3.0]
x = pd.DataFrame([
[ 0.,0.,0.],
[ 0.32, -6.79, 0.],
[ 0.,0.,0.],
[ 0.,1.5,0.],
[ 0.,0., -1.71]])
b = pd.DataFrame(myPosFunc(x[x>0], I)) #calculate all positive values
c = pd.DataFrame(myNegFunc(x[x<0], I)) #calculate all negative values
b = b.combineMult(c) #put values of c in b
b = b.fillna(1) #replace all missing values that were '0' in the raw array
y = b.product() #multiply all elements in one row
#Output
print ('final result')
print (y)
print (y.tolist())