Numpy has a function for this exact problem: numpy.linalg.solve
To construct the matrix we first need to digest the string turning it into an array of coefficients and solutions.
Finding Numbers
First we need to write a function that takes a string like "c_1 3" and returns the number 3.0. Depending on the format you want in your input string you can either iterate over all chars in this array and stop when you find a non-digit character, or you can simply split on the space and parse the second string. Here are both solutions:
def find_number(sub_expr):
"""
Finds the number from the format
number*string or numberstring.
Example:
3x -> 3
4*x -> 4
"""
num_str = str()
for char in sub_expr:
if char.isdigit():
num_str += char
else:
break
return float(num_str)
or the simpler solution
def find_number(sub_expr):
"""
Returns the number from the format "string number"
"""
return float(sub_expr.split()[1])
Note: See edits
Get matrices
Now we can use that to split each expression into two parts: The solution and the equation by the "=". The equation is then split into sub_expressions by the "+" This way we would end turn the string "3x+4y = 3" into
sub_expressions = ["3x", "4y"]
solution_string = "3"
Each sub expression then needs to be fed into our find_numbers function. The End result can be appended to the coefficient and solution matrices:
def get_matrices(expressions):
"""
Returns coefficient_matrix and solutions from array of string-expressions.
"""
coefficient_matrix = list()
solutions = list()
last_len = -1
for expression in expressions:
# Note: In this solution all coefficients must be explicitely noted and must always be in the same order.
# Could be solved with dicts but is probably overengineered.
if not "=" in expression:
print(f"Invalid expression {expression}. Missing "="")
return False
try:
c_string, s_string = expression.split("=")
c_strings = c_string.split("+")
solutions.append(float(s_string))
current_len = len(c_strings)
if last_len != -1 and current_len != last_len:
print(f"The expression {expression} has a mismatching number of coefficients")
return False
last_len = current_len
coefficients = list()
for c_string in c_strings:
coefficients.append(find_number(c_string))
coefficient_matrix.append(coefficients)
except Exception as e:
print(f"An unexpected Runtime Error occured at {coefficient}")
print(e)
exit()
return coefficient_matrix, solutions
Now let's write a simple main function to test this code:
# This is not the code you want to copy-paste
# Look further down.
from sys import argv as args
def main():
expressions = args[1:]
matrix, solutions = get_matrices(expressions)
for row in matrix:
print(row)
print("")
print(solutions)
if __name__ == "__main__":
main()
Let's run the program in the console!
user:$ python3 solve.py 2x+3y=4 3x+3y=2
[2.0, 3.0]
[3.0, 3.0]
[4.0, 2.0]
You can see that the program identified all our numbers correctly
AGAIN: use the find_number function appropriate for your format
Put The Pieces Together
These Matrices now just need to be pumped directly into the numpy function:
# This is the main you want
from sys import argv as args
from numpy.linalg import solve as solve_linalg
def main():
expressions = args[1:]
matrix, solutions = get_matrices(expressions)
coefficients = solve_linalg(matrix, solutions)
print(coefficients)
# This bit needs to be at the very bottom of your code to load all functions first.
# You could just paste the main-code here, but this is considered best-practice
if __name__ == '__main__':
main()
Now let's test that:
$ python3 solve.py x*2+y*4+z*0=20 x*1+y*1+z*-1=3 x*2+y*2+z*-3=3
[2. 4. 3.]
As you can see the program now solves the functions for us.
Out of curiosity: Math homework? This feels like math homework.
Edit: Had a typo "c_string" instead of "c_strings" worked out in all tests out of pure and utter luck.
Edit 2: Upon further inspection I would reccomend to split the sub-expressions by a "*":
def find_number(sub_expr):
"""
Returns the number from the format "string number"
"""
return float(sub_expr.split("*")[1])
This results in fairly readable input strings
