Chap_7-2-2-ans.ipynb

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    "# 7の課題"
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    "学籍番号："
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   "source": [
    "氏　　名："
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   "source": [
    "import sympy as sp\n",
    "\n",
    "pi_val = sp.N(sp.pi, 100000000)\n",
    "pi_lis = [int(i) for i in str(pi_val).replace('.', '')]"
   ]
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   "source": [
    "□2-2. 円周率$\\pi$の小数第99999999位まである１桁の数の個数（すべての数の個数は100000000個）を，長さ10（10個の要素）のリスト`cnt_lis`に求めたい。つまり，`cnt_lis`の，インデックス0には0の個数，インデックス1には1の個数，インデックス2には2の個数，...，インデックス9には9の個数が入る。以下のコードの３行目をコーディングして実行し，`cnt_lis`の内容を表示せよ。"
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      "[9999922, 10002475, 10001091, 9998443, 10003863, 9993478, 9999417, 9999610, 10002180, 9999521]\n",
      "CPU times: user 2.9 s, sys: 1.15 ms, total: 2.9 s\n",
      "Wall time: 2.9 s\n"
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    "%%time\n",
    "cnt_lis = [0]*10\n",
    "for k in pi_lis:\n",
    "    cnt_lis[k] += 1   #この行をコーディングせよ\n",
    "print(cnt_lis)"
   ]
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   "source": [
    "*□2-2の別解. 下記のように`count`関数を用いると`for`ループが10回で済む（2-2は100000000回のループ）ので，上のコードより圧倒的に速い（と考えたが，結果では上の方が速い...countメソッドがボトルネックか？）。ただし，Chapter 7までに`range`関数が出てこないのでここではこの解答は避けることにする。*"
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      "[9999922, 10002475, 10001091, 9998443, 10003863, 9993478, 9999417, 9999610, 10002180, 9999521]\n",
      "CPU times: user 4.45 s, sys: 4.15 ms, total: 4.46 s\n",
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    "%%time\n",
    "cnt_lis1 = [0]*10\n",
    "for k in range(10):\n",
    "    cnt_lis1[k] = pi_lis.count(k)\n",
    "print(cnt_lis1)"
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