2020-06-20 課題

メモ：先に進む前に録画してあるか確認しよう

自分用メモ

Matplotlib

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import numpy as np
import matplotlib.pyplot as plt

x = np.linspace(-4, 4, 201)
y1 = x**2 - 2 * x + 1
y2 = x**3 + 2 * x + 3
y3 = x**4 - x**2 - 2 * x + 3

plt.plot(x, y1, label="y1")
plt.plot(x, y2, label="y2")
plt.plot(x, y3, label="y3")

plt.grid()
plt.legend()
#plt.axes().set_aspect('equal', 'datalim') # アスペクト比を合わせる
plt.show()

自然言語処理100本ノック

第1章: 準備運動 00. 文字列の逆順

文字列"stressed"の文字を逆に（末尾から先頭に向かって）並べた文字列を得よ．

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print("".join(list(reversed("stressed"))))
print("stressed"[::-1])

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desserts
desserts

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reversed("stressed")

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<reversed at 0x7fddcfffd5f8>

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"".join(reversed("stressed"))

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'desserts'

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list(reversed("stressed"))

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['d', 'e', 's', 's', 'e', 'r', 't', 's']

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"|".join(list(reversed("stressed")))

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'd|e|s|s|e|r|t|s'

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"stressed"[::2]

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'srse'

第1章: 準備運動 01. 「パタトクカシーー」

  • 「パタトクカシーー」という文字列の1,3,5,7文字目を取り出して連結した文字列を得よ．
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print("パタトクカシーー"[1::2])

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タクシー

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print("パタトクカシーー"[0::2])

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パトカー

02. 「パトカー」＋「タクシー」＝「パタトクカシーー」

「パトカー」＋「タクシー」の文字を先頭から交互に連結して文字列「パタトクカシーー」を得よ．

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s1 = "パトカー"
s2 = "タクシー"


print("".join([s1[i] + s2[i] for i in range(4)]))


s = ""
for i in range(4):
    print(s)
    s = s + s1[i]+ s2[i]
print(s)

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パタトクカシーー

パタ
パタトク
パタトクカシ
パタトクカシーー

03 円周率

"Now I need a drink, alcoholic of course, after the heavy lectures involving quantum mechanics."という文を単語に分解し，各単語の（アルファベットの）文字数を先頭から出現順に並べたリストを作成せよ．

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s = "Now I need a drink, alcoholic of course, after the heavy lectures involving quantum mechanics."
print(list(map(lambda x: len(x), s.split())))

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[3, 1, 4, 1, 6, 9, 2, 7, 5, 3, 5, 8, 9, 7, 9]

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s = "Now I need a drink, alcoholic of course, after the heavy lectures involving quantum mechanics."
s.split()

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['Now',
 'I',
 'need',
 'a',
 'drink,',
 'alcoholic',
 'of',
 'course,',
 'after',
 'the',
 'heavy',
 'lectures',
 'involving',
 'quantum',
 'mechanics.']

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s = "Now I need a drink, alcoholic of course, after the heavy lectures involving quantum mechanics."
print(list(map(lambda x: len(x), s.replace(",", "").replace(".", "").split())))

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[3, 1, 4, 1, 5, 9, 2, 6, 5, 3, 5, 8, 9, 7, 9]

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s = "Now I need a drink, alcoholic of course, after the heavy lectures involving quantum mechanics."
print(s.replace(",", ""))
print(s.replace(",", "").replace(".", ""))
print(s.replace(",", "").replace(".", "").split())

xs = []
for x in s.replace(",", "").replace(".", "").split():
    print(len(x))
    xs.append(len(x))
print(xs)

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Now I need a drink alcoholic of course after the heavy lectures involving quantum mechanics.
Now I need a drink alcoholic of course after the heavy lectures involving quantum mechanics
['Now', 'I', 'need', 'a', 'drink', 'alcoholic', 'of', 'course', 'after', 'the', 'heavy', 'lectures', 'involving', 'quantum', 'mechanics']
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1
4
1
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9
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3
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[3, 1, 4, 1, 5, 9, 2, 6, 5, 3, 5, 8, 9, 7, 9]

04. 元素記号

"Hi He Lied Because Boron Could Not Oxidize Fluorine. New Nations Might Also Sign Peace Security Clause. Arthur King Can."という文を単語に分解し，1, 5, 6, 7, 8, 9, 15, 16, 19番目の単語は先頭の1文字，それ以外の単語は先頭の2文字を取り出し，取り出した文字列から単語の位置（先頭から何番目の単語か）への連想配列（辞書型もしくはマップ型）を作成せよ．

（言っていることがよくわからなくて何度も読み返した。）

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numbers = [1,5,6,7,8,9,15,16,19]
numbers = list(map(lambda x: x - 1, numbers))

def f(i,v):
    return v[0] if i in numbers else v[0:2]

s = "Hi He Lied Because Boron Could Not Oxidize Fluorine. New Nations Might Also Sign Peace Security Clause. Arthur King Can."
s = s.replace(",", "").replace(".", "").split()
print({f(i,v): i for i, v in enumerate(s)})

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{'H': 0, 'He': 1, 'Li': 2, 'Be': 3, 'B': 4, 'C': 5, 'N': 6, 'O': 7, 'F': 8, 'Ne': 9, 'Na': 10, 'Mi': 11, 'Al': 12, 'Si': 13, 'P': 14, 'S': 15, 'Cl': 16, 'Ar': 17, 'K': 18, 'Ca': 19}

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s = ""
if i in numbers:
    s = v[0]
else:
    s = v[0:2]

内包表記をループで書き直した

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numbers = [1,5,6,7,8,9,15,16,19]
numbers = list(map(lambda x: x - 1, numbers))

def f(i,v):
    return v[0] if i in numbers else v[0:2]

s = "Hi He Lied Because Boron Could Not Oxidize Fluorine. New Nations Might Also Sign Peace Security Clause. Arthur King Can."
s = s.replace(",", "").replace(".", "").split()

dic = {}
for i,v in enumerate(s):
    dic[f(i,v)] = i

print(dic)

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0
Hi

1
He

2
Lied

3
Because

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Boron

5
Could

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Not

7
Oxidize

8
Fluorine

9
New

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Nations

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Might

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Also

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Sign

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Peace

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Security

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Clause

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Arthur

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King

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Can

{'H': 0, 'He': 1, 'Li': 2, 'Be': 3, 'B': 4, 'C': 5, 'N': 6, 'O': 7, 'F': 8, 'Ne': 9, 'Na': 10, 'Mi': 11, 'Al': 12, 'Si': 13, 'P': 14, 'S': 15, 'Cl': 16, 'Ar': 17, 'K': 18, 'Ca': 19}

プログラミングの一般論

web システムの事例

データ構造とアルゴリズム

（連結）リストと配列

リストの特徴

配列の特徴

ベクター（参考）