如何使用正则表达式找到方程中的截距项

Question

这里我的代码可以为我获取x和y的系数，但我没有得到截距项！

eq = ["1x+1y+2", "-1x+12Y-6", "2-5y-3x", "7y-50+2X", "3.14x-1.5y+9", "11.0x-1.5y+9.8"]
def coefficients(equation):
import re
coef_x = re.findall('-?[0-9.]*[Xx]', equation)[0][:-1]
coef_y = re.findall('-?[0-9.]*[Yy]', equation)[0][:-1]
intercept = re.findall("b[+-]?d+[.]?[d]+b|[+-]?d+[+-]|[+-]?d+[s]", equation)

#return float(coef_x), float(coef_y), intercept
print(coef_x, coef_y, intercept)
print("")
for i in eq:
coefficients(i)

输出：

1 1 []
-1 12 []
-3 -5 ['2-']
2 7 ['-50+']
3.14 -1.5 []
11.0 -1.5 []

预期：

1 1 2
-1 12 -6
-3 -5 2
2 7 -50
3.14 -1.5 9
11.0 -1.5 9.8

如果有人能提出必要的改变，我会非常感激，这样我就可以得到正确的拦截

Answer 1

试试这个：

import re
eq = ["1x+1y+2", "-1x+12Y-6", "2-5y-3x", "7y-50+2X", "3.14x-1.5y+9", "11.0x-1.5y+9.8"]
x = re.compile("([+-]?d+(.d+)?)[xX]")
y = re.compile("([+-]?d+(.d+)?)[yY]")
z = re.compile("([+-]?d+(.d+)?)([+-]|$)")
for line in eq:
xx = 0
m = x.search(line)
if m:
xx = m.group(1)
yy = 0
m = y.search(line)
if m:
yy = m.group(1)
zz = 0
m = z.search(line)
if m:
zz = m.group(1)
print("X = "+ xx + ", Y = " + yy + ", intercept= " + zz)

Answer 2

实际上，您可以使用前瞻断言使用一个正则表达式来查找系数，这不会推进正则表达式引擎的扫描位置。

(?i)(?=.*?(-?[d.]+(?=x)))(?=.*?(-?[d.]+(?=y)))(?=.*?(-?[d.]+(?=(?:+|-|$))))

参见Regex Demo

例如，要查找拦截：

(?=                   # start of a positive lookahead assertion
.*?              # match 0 or more characters non-greedily
(                # start of capture group 3
-?          # match optional "-"
[d.]+       # match number (not very sophisticated)
(?=          # start of a new positive lookahead assertion that must be met
(?:      # start of a non-capture group
+   # match "+"
|    # or
-   # match "-"
|    # or
$   # end of string
)
)
)
)

本质上，我们正在寻找一个后面跟着"+"、"-"或字符串末尾的数字。同样，为了找到x的系数，我们寻找一个后面跟着"x"的数字：

(?=                   # start of a positive lookahead assertion
.*?              # match 0 or more characters non-greedily
(                # start of capture group 1
-?           # match optional "-"
[d.]+        # match number (not very sophisticated)
(?=           # start of a new positive lookahead assertion that must be met
x         # match "x"
)
)
)

代码：

import re
eq = ["1x+1y+2", "-1x+12Y-6", "2-5y-3x", "7y-50+2X", "3.14x-1.5y+9", "11.0x-1.5y+9.8"]
regex = re.compile(r'(?i)(?=.*?(-?[d.]+(?=x)))(?=.*?(-?[d.]+(?=y)))(?=.*?(-?[d.]+(?=(?:+|-|$))))')
for ex in eq:
m = regex.search(ex)
if m:
print(m.group(1), m.group(2), m.group(3))

打印：

1 1 2
-1 12 -6
-3 -5 2
2 7 -50
3.14 -1.5 9
11.0 -1.5 9.8

Answer 3

同一问题的新解决方案。

eq = ["1x+1y+2", "-1x+12Y-6", "2-5y-3x", "7y-50+2X", "3.14x-1.5y+9", "11.0x-1.5y+9.8"]
def coefficients(equation):
import re
coef_x = re.findall('-?[0-9.]*[Xx]', equation)[0][:-1]
coef_y = re.findall('-?[0-9.]*[Yy]', equation)[0][:-1]
intercept = re.sub("[+-]?d+[XxYy]|[+-]?d+.d+[XxYy]","", equation)

return float(coef_x), float(coef_y), float(intercept)

for i in eq:
print(coefficients(i))

输出：

(1.0, 1.0, 2.0)
(-1.0, 12.0, -6.0)
(-3.0, -5.0, 2.0)
(2.0, 7.0, -50.0)
(3.14, -1.5, 9.0)
(11.0, -1.5, 9.8)