Задачи по Python и машинному обучению. Страница 17

s1 = 'a\u00f1'
s2 = 'an\u0303'
print(s1)  # añ
print(s2)  # añ
print(s1 == s2)  # False

import re
s = "a x \n x a"
p1 = re.compile(r"a.*?a")
p2 = re.compile(r"a.*?a", re.DOTALL)
r1 = p1.findall(s)
r2 = p2.findall(s)
print(len(r1), len(r2))

import re
s = "<span>1</span><b>2</b>"
p1 = re.compile(r"(<\w+>.*</\w+>)")
p2 = re.compile(r"(<\w+>.*?</\w+>)")
x1 = p1.findall(s)
x2 = p2.findall(s)
print(len(x1), len(x2))

import re
expr = "a + A"
p = re.compile(r'a', flags=re.IGNORECASE)
g = lambda t, mt: t.upper() if mt.isupper() else t.lower()
f = lambda text: lambda m: g(text, m.group())
r = p.sub(f('b'), expr)
print(r)

import re
expr = 'x = 3; y = x**2; print(y)'
p = re.compile(r'\*\*(\d+)')
f = lambda m: f'^{int(m.group(1)) + 1}'
r = p.sub(f, expr)
exec(r)

import re
p = re.compile(r'(\d+)\*(\d+)')
expr = "2*3 + 4*5"
x, n = p.subn(r'\2 - \1', expr)
print(eval(x), n)

import re
p = re.compile(r'(\d+)\+(\d+)\-(\d+)')
expr = '2+10-5'
x = p.sub(r'\1*\2/\3', expr)
print(eval(x))

import re
d = "x\d"              
t = "xx97"

import re
d = re.compile('(\d+)/(\d+)')
text = '05/06 and 07/08'
x = d.match(text).group(0)
y = d.match(text).groups()[0]
print(x == y)

from sklearn.linear_model import LinearRegression
import numpy as np
y = np.array([0, 2, 4])
n = len(y)
X = np.arange(n).reshape((n, 1))
model = LinearRegression()
model.fit(X, y)
p = model.predict([[3], [4]])
print(*(round(x) for x in p))

from fnmatch import fnmatchcase
l = [f'{b}{a}' for a in range(20) for b in ['x', 'y']]
r = [x for x in l if fnmatchcase(x, 'x*0')]
print(*r)

from fnmatch import fnmatch
n = "foo2022.txt"
x = fnmatch(n, "foo[0-9]*")
y = fnmatch(n, "foo[0-9]+")
print(x, y)

import re
line = "a+b"
x = re.split(r"(\+)", line)
y = re.split(r"\+", line)
z = re.split(r"(?:\+)", line)
p = set(x) - set(y) | set(z)
print(len(p))