Yet another "simple" headscratcher

J

Josh English

I am trying to whip up a quick matrix class that can handle multiplication.

Should be no problem, except when it fails.

--- Begin
#!/usr/bin/env python
# _*_ coding: utf-8

from operator import mul

class Matrix(object):
"""Matrix([data])
Data should be a list of equal sized lists.
Defaults to a 2d identity matrix
"""
def __init__(self, data=None):
if data is None:
data = [[1,0], [0,1]]

self.data = []
if isinstance(data, (list, tuple)):
ncols = len(data[0])
for row in data:
if len(row) != ncols:
raise ValueError("Rows are unequal lengths")
self.data.append(row)
self.size = (len(self.data), len(self.data[0]))

def get_row(self, idx):
if (0 <= idx < self.size[0]):
return self.data[idx]
else:
raise ValueError("Bad row")

def get_col(self, idx):
if (0 <= idx < self.size[1]):
return list(d[idx] for d in self.data)
else:
raise ValueError("Bad column index")

def __mul__(self, other):
if not isinstance(other, (Matrix,int, float)):
raise ValueError("Cannot multiply by given value")
if isinstance(other, (int, float)):
res = []
for row in self.data:
res.append([d*other for d in row])
return Matrix(res)
# left with a matrix
res = zero_matrix(self.size[0], other.size[1])
for i in range(res.size[0]):
for j in range(res.size[1]):
print i, j, self.get_row(i), other.get_col(j),
temp = map(mul, self.get_row(i), other.get_col(j))

print temp,
t = sum(temp)
print t
res.data[j] = t
print res.data
return res

def as_string(self):
# return a list of lines that look pretty
stringed =[]
for row in self.data:
stringed.append(map(str, row))
widths = []
for col in range(self.size[1]):
column = [s[col] for s in stringed]
widths.append(max(map(len, column)))
item_format = "{:>%s}"
format_items = [item_format % w for w in widths]
format_string = " ".join(format_items)

formatted = [format_string.format(*s) for s in stringed]
return formatted

def zero_matrix(rows, cols):
row = [0] * cols
data = []
for r in range(rows):
data.append(row)

return Matrix(data)

M = Matrix(([1, 0], [0, -1]))

N = M*4


print '\n'.join(M.as_string())
print '-'
print '\n'.join(N.as_string())


print '-'
S = N * M
print '\n'.join(S.as_string())
--- END

For some reason, my output from this is:

1 0
0 -1
-
4 0
0 -4
-
0 0 [4, 0] [1, 0] [4, 0] 4
[[4, 0], [4, 0]]
0 1 [4, 0] [0, -1] [0, 0] 0
[[4, 0], [4, 0]]
1 0 [0, -4] [1, 0] [0, 0] 0
[[0, 0], [0, 0]]
1 1 [0, -4] [0, -1] [0, 4] 4
[[0, 4], [0, 4]]
0 4
0 4
The print lines prove to me that the logic is working, but for some reason, assigning the sum to a particular item in a particular row is assigning the same row values to every row.

This should be one of those really simple Python things, but for the life of me I don't see it.

The first [[4, 0], [4, 0]] is clearly wrong. In each step, this algorithm is repeating the row.

Any ideas as to why this is happening?

Python 2.7.5, Windows 7, so nothing exotic.

Josh
 
I

Ian Kelly

J

Josh English

Mea culpa, gang.

I found it.

It had absolutely nothing to do with the multiplication.

It was in zero_matrix.

I feel like a fool.

Josh
 
G

Gary Herron

...

def zero_matrix(rows, cols):
row = [0] * cols
data = []
for r in range(rows):
data.append(row)

return Matrix(data)

There is a simple and common newbie mistake here. It looks like you
are appending several copies of a zero row to data, but in fact you are
appending multiple references to a single row. (The hint is that you
only created *one* row.)

Put the
row = [0] * cols
inside the loop so each append is using its own row rather than one
shared row being used multiple times.


Here's a small example that demonstrates problem:
row = [0,0,0,0]
data = []
data.append(row)
data.append(row)
data[0][0] = 99
data
[[99, 0, 0, 0], [99, 0, 0, 0]]


Gary Herron
 
W

wxjmfauth

# from my lib.... val = float(val)
.... return [[val] * nc for i in range(nr)]
....( 0.00000e+000 0.00000e+000 0.00000e+000 )
( 0.00000e+000 0.00000e+000 0.00000e+000 )
( 3.14160e+000 0.00000e+000 0.00000e+000 )
( 0.00000e+000 0.00000e+000 0.00000e+000 )
( 3.14160e+000 0.00000e+000 0.00000e+000 )
( 0.00000e+000 1.23450e+000 0.00000e+000 )
jmf
 

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