& Emsp; Hello, this is mana binding. This is an article that I will introduce because I was able to make a program that I made a lot because of the necessity. I made a program to solve a system of linear equations that is reasonably practical. Please kindly let me know if there are any bugs. Please note that we will not explain how to sweep out.
The & emsp; environment is Python3 series.
> python --version
Python 3.9.0
The & emsp; program is shown below.
solve_PE.py
import sys
def hakidashi(m, show_matrix):
num_line = len(m)
num_column = len(m[0])
div = lambda a, b: 0 if b==0 else a / b
for p in range(num_line):
m[p] = [div(n, m[p][p]) for n in m[p]]
for i in range(num_line):
x = div(m[i][p], m[p][p])
for j in range(num_column):
if(i != p):
m[i][j] = m[i][j] - x * m[p][j]
if(show_matrix):
for e in m:
print(e)
return list(map(lambda m: m[-1], m))
def dec_to_frac(f, max_denominator, calc_accuracy):
for i in range(1, max_denominator + 1):
k1 = int(f * i)
k2 = k1 + 1
if(abs((k1 / i) - f) < calc_accuracy):
return f"{k1}/{i}"
elif(abs((k2 / i) - f) < calc_accuracy):
return f"{k2}/{i}"
else:
return "not found"
def read_file(file_name, file_codec):
c, m = [], []
with open(file_name, encoding=file_codec) as f:
lists = list(f.readlines())
c = split_line(lists[0])
for l in lists[1:]:
m.append(split_line(l, num=True))
return c, m
def split_line(line, num=False):
line = line.strip()
line += "#"
data_line = []
word = ""
is_firstblank = True
for e in line:
if(" " == e and is_firstblank):
data_line.append(word)
is_firstblank = False
word = ""
elif("," == e):
data_line.append(word)
word = ""
elif("#" == e):
data_line.append(word)
break
else:
is_firstblank = True
word += e
else:
data_line.append(word)
return [
float(x.strip())
if num else
x.strip()
for x in data_line
if x not in ["", " "]
]
if __name__ == "__main__":
file_name = sys.argv[1]
file_encode = "utf-8"
max_denominator = 10000000
calc_accuracy = 0.0000000001
show_matrix = True
c, m = read_file(file_name, file_encode)
if(len(c) + 1 == len(m[0])):
ans = hakidashi(m, show_matrix)
for c, ans in zip(c, ans):
f = dec_to_frac(ans, max_denominator, calc_accuracy)
print(f"{c} = {ans} ({f})")
else:
print("data error", file=sys.stderr)
& emsp; Below is a description of each.
& emsp; hakidashi () is the corresponding function. There is no particular ingenuity, but in order to avoid the error of dividing by zero, a function called div is defined internally, and 0 is returned when it seems to be divided by zero. You can judge that print is done just before the return because the constant term does not become 1 correctly when an equation that has no solution is passed.
& emsp; dec_to_frac () is the corresponding function. It verifies while changing the denominator, and returns a character string expressed as a fraction when the precision is acceptable. If it is not found in the range, not found is displayed. This is a for loop and the number of calculations is large, so if it is too heavy, adjust the values of max_denominator and calc_accuracy.
I referred to this site. FloatToFraction (JavaScript version)
& emsp; read_file () and split_line () are the corresponding functions.
& emsp; read_file opens the file data from the file name given from the command line and passes it to split_line. c is an array that contains character labels (x, y, etc.), and m is numeric array data.
& emsp; split_line is a polite division so that it can be read in comma separate or space separate, or comments can be written.
For example, use a file with the following text:
mat1.txt
x y z
4 2 1 26
1 6 3 48
4 1 5 34
The above represents the following simultaneous equations.
> python solve_PE.py mat1.txt
[1.0, 0.0, 0.0, 2.727272727272727]
[0.0, 1.0, 0.0, 5.818181818181818]
[0.0, 0.0, 1.0, 3.454545454545455]
x = 2.727272727272727 (30/11)
y = 5.818181818181818 (64/11)
z = 3.454545454545455 (38/11)
Also, consider the case where there is no solution as shown below.
mat2.txt
x1, x2 #The lower formula is the upper formula
3, 6, 9 #Because it's just doubled
6, 12, 18 #There is no solution
In addition, # represents a comment, and it is possible to actually describe it in a file.
> python solve_PE.py mat2.txt
[1.0, 2.0, 3.0]
[0, 0, 0]
x1 = 3.0 (3/1)
x2 = 0 (0/1)
The values are entered appropriately, but you can see that it is strange because the matrix above is not triangular.
& emsp; Thank you for your hard work. Did it work? It's a program that can be used quite well even though it's less than a hundred lines, so I think Python is amazing, and the algorithm that my predecessor thought was amazing (impression of elementary school students). It's fun to be able to build it yourself, and I want to live with happiness in these things. & emsp; Thank you for reading to the end. I hope it helps. It doesn't matter, but I can't really count on writing something next. I am surprised to see past articles by myself. I've got my hands on it, so I'll put it around here. Thank you very much.
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