def main():
text = "paraparaparadise"
x = get_n_gram.n_gram(list(text))
print(x)
difx = dif(x)
print(difx)
text = "paragraph"
y = get_n_gram.n_gram(list(text))
print(y)
dify = dif(y)
print(dify)
union = (difx | dify)
intersection = (difx & dify)
difference = (difx - dify)
print(union)
print(intersection)
print(difference)
if "se" in difx:
print(u"found_se_in_x")
if "se" in dify:
print(u"found_se_in_y")
def main():
text = "paraparaparadise"
x = get_n_gram.n_gram(list(text))
print(x)
difx = dif(x)
print(difx)
text = "paragraph"
y = get_n_gram.n_gram(list(text))
print(y)
dify = dif(y)
print(dify)
union = (difx | dify)
intersection = (difx & dify)
difference = (difx - dify)
print(union)
print(intersection)
print(difference)
if "se" in difx:
print (u"found_se_in_x")
if "se" in dify:
print (u"found_se_in_y")
# -*- encoding: utf-8 -*-
from knock05 import n_gram
X = set(n_gram("paraparaparadise", 2))
Y = set(n_gram("paragraph", 2))
print("X " + str(X))
print("Y " + str(Y))
print("和集合 " + str(X.union(Y)))
print("積集合 " + str(X.intersection(Y)))
print("X - Y " + str(X.difference(Y)))
print("Y - X " + str(Y.difference(X)))
print("seがXに含まれるか？" + str("se" in X))
print("seがYに含まれるか？" + str("se" in Y))
"""
import os
import sys
from knock05 import n_gram
sys.path.append(os.path.join(os.path.dirname(__file__), "../../"))
from kiyuna.utils.message import Renderer, message # noqa: E402 isort:skip
if __name__ == "__main__":
w1 = "paraparaparadise"
w2 = "paragraph"
n = 2
X = set(n_gram(w1, n))
Y = set(n_gram(w2, n))
tgt = n_gram("se", n).pop()
print("X =", X)
print("Y =", Y)
print("X ∪ Y = {}".format(X | Y)) # X.union(Y)
print("X ∩ Y = {}".format(X & Y)) # X.intersection(Y)
print("X \\ Y = {}".format(X - Y)) # X.difference(Y)
print("Y \\ X = {}".format(Y - X)) # Y.difference(X)
print(f"X includes 'se': {tgt in X}")
print(f"Y includes 'se': {tgt in Y}")
with Renderer("MEMO") as out:
'''
06. 集合
"paraparaparadise" と "paragraph" に含まれる文字 bi-gram の集合を，
それぞれ, X と Y として求め，X と Y の和集合，積集合，差集合を求めよ．
さらに，'se' という bi-gram が X および Y に含まれるかどうかを調べよ．
'''
from knock05 import n_gram # -B をつけるとキャッシュが出ない
if __name__ == '__main__':
w1 = "paraparaparadise"
w2 = "paragraph"
tgt = n_gram('se', n=2).pop()
X = set(n_gram(w1, n=2))
Y = set(n_gram(w2, n=2))
print('X =', X)
print('Y =', Y)
print('X ∪ Y = {}'.format(X | Y)) # X.union(Y)
print('X ∩ Y = {}'.format(X & Y)) # X.intersection(Y)
print('X \ Y = {}'.format(X - Y)) # X.difference(Y)
print('Y \ X = {}'.format(Y - X)) # Y.difference(X)
print(f"X includes 'se': {tgt in X}")
print(f"Y includes 'se': {tgt in Y}")
from knock05 import n_gram
str1 = "paraparaparadise"
str2 = "paragraph"
n = 2
X = set(n_gram(str1, n))
Y = set(n_gram(str2, n))
print(f"{X | Y}")
print(f"{X & Y}")
print(f"{X - Y}")
if "se" in X:
print("Xに含まれる")
else:
print("Xに含まれない")
if "se" in Y:
print("Yに含まれる")
else:
print("Yに含まれない")
# 06. 集合
# “paraparaparadise”と”paragraph”に含まれる文字bi-gramの集合を，それぞれ, XとYとして求め，XとYの和集合，積集合，差集合を求めよ．
# さらに，’se’というbi-gramがXおよびYに含まれるかどうかを調べよ．
from knock05 import n_gram
if __name__ == "__main__":
str1 = "paraparaparadise"
str2 = "paragraph"
str3 = "se"
X = n_gram(str1, t="char")
X = set(X)
Y = n_gram(str2, t="char")
Y = set(Y)
union = X | Y
intersec = X & Y
# 差集合演算は可換でないので、「XとYの差集合」という表現はよくない。
diff1 = X - Y
diff2 = Y - X
print(f"{X=}")
print(f"{Y=}")
print(f"{union=}")
print(f"{intersec=}")
print(f"{diff1=}")
print(f"{diff2=}")
# 「XおよびYに含まれるかどうかを調べよ」は、「XとYのそれぞれに対して別々に調べる」という意味だとは思うが、「XとYの両方に含まれるかどうかを調べる」という意味にも解釈できる？
# 「および」は、文脈によって、「かつ」にも「または」にも言い換えられる？
ans1 = str3 in X
ans2 = str3 in Y
ans3 = str3 in intersec
print(f"{str3}はXに含まれるか：{ans1}")
from knock05 import n_gram
s1 = "paraparaparadise"
s2 = "paragraph"
bg1 = n_gram(2, s1)
bg2 = n_gram(2, s2)
se = n_gram(2, "se")[0]
print("Union:", set(bg1) or set(bg2))
print("Intersection:", set(bg1) and set(bg2))
print("minus:", set(bg1) - set(bg2))
print(se, "in bigram(", s1, "):", se in bg1)
print(se, "in bigram(", s2, "):", se in bg2)
from knock05 import n_gram
s1 = "paraparaparadise"
s2 = "paragraph"
X = set(n_gram(s1, 2)[0])
Y = set(n_gram(s2, 2)[0])
print(f"和集合 : {X | Y}")
print(f"積集合 : {X & Y}")
print(f"差集合 : {X - Y}")
if "se" in X:
print("'se' is in X.")
else:
print("'se' is not in X.")
if "se" in Y:
print("'se' is in Y.")
else:
print("'se' is not in Y.")
import knock05
s1 = 'paraparaparadise'
s2 = 'paragraph'
X = set(knock05.n_gram(2, s1, knock05.CHAR))
Y = set(knock05.n_gram(2, s2, knock05.CHAR))
print(f'{"X":<15}: {X}')
print(f'{"Y":<15}: {Y}')
print(f'{"Union":<15}: {X | Y}')
print(f'{"Intersection":<15}: {X & Y}')
print(f'{"Difference(X-Y)":<15}: {X - Y}')
print(f'{"Difference(Y-X)":<15}: {Y - X}')
print(f"Is 'se' belong to X? -> {'se' in X}")
print(f"Is 'se' belong to Y? -> {'se' in Y}")
# 06. 集合
# "paraparaparadise"と"paragraph"に含まれる文字bi - gramの集合を，それぞれ, XとYとして求め，XとYの和集合，積集合，差集合を求めよ．
# さらに，'se'というbi - gramがXおよびYに含まれるかどうかを調べよ．
# 実行結果
# X : {'is', 'pa', 'ra', 'ad', 'ar', 'se', 'ap', 'di'}
# Y : {'gr', 'ag', 'ph', 'pa', 'ra', 'ar', 'ap'}
# Union : {'gr', 'ag', 'ph', 'is', 'pa', 'ra', 'ad', 'ar', 'se', 'ap', 'di'}
# Intersection : {'ra', 'ar', 'ap', 'pa'}
# Difference(X-Y): {'ad', 'is', 'di', 'se'}
# Difference(Y-X): {'ag', 'gr', 'ph'}
# Is 'se' belong to X? -> True
# Is 'se' belong to Y? -> False
# 06. 集合
# "paraparaparadise"と"paragraph"に含まれる文字bi-gramの集合を，
# それぞれ, XとYとして求め，XとYの和集合，積集合，差集合を求めよ．
# さらに，'se'というbi-gramがXおよびYに含まれるかどうかを調べよ．
from knock05 import n_gram
if __name__ == "__main__":
target1 = "paraparaparadise"
target2 = "paragraph"
# n_gram 関数はリストで返されるので、集合演算を行うために set 型にキャスト
x = set(n_gram(target1, 2))
y = set(n_gram(target2, 2))
print(f"X : {x}")
print(f"Y : {y}")
print(f"union : {x | y}")
print(f"intersection : {x & y}")
print(f"difference(X-Y) : {x - y}")
print(f"difference(Y-X) : {y - x}")
print(f"'se' in X? : {'se' in x}")
print(f"'se' in Y? : {'se' in y}")
from knock05 import n_gram
X = set(n_gram("paraparaparadise"))
Y = set(n_gram("paragraph"))
print("X or Y: {}".format(X | Y))
print("X and Y: {}".format(X & Y))
print("X - Y: {}".format(X - Y))
print("Y - X: {}".format(Y - X))
print("'se' in X?: {}".format("se" in X))
print("'se' in Y?: {}".format("se" in Y))
from knock05 import n_gram
'''
"paraparaparadise"と"paragraph"に含まれる文字bi-gramの集合を，それぞれ, XとYとして求め，XとYの和集合，積集合，差集合を求めよ．
さらに，'se'というbi-gramがXおよびYに含まれるかどうかを調べよ．
'''
if __name__ == '__main__':
s1 = "paraparaparadise"
s2 = "paragraph"
X = set(n_gram(s1, 2))
Y = set(n_gram(s2, 2))
print(X.union(Y), X.intersection(Y), X.difference(Y), Y.difference(X))
if 'se' in X.intersection(Y):
print('Yes')
else:
print('No')
