{"type":"rich","height":"190","categories":[],"title":"\u52dd\u624b\u306b\u6dfb\u524a (C vs Python vs Ruby vs Haskell\uff08\u7121\u610f\u5473\u306a\u51e6\u7406de\u30d9\u30f3\u30c1\u30de\u30fc\u30af\uff09)","description":"\u306a\u3093\u304b\u6307\u540d\u3055\u308c\u305f\u306e\u3067\u3001 C vs Python vs Ruby vs Haskell\uff08\u7121\u610f\u5473\u306a\u51e6\u7406de\u30d9\u30f3\u30c1\u30de\u30fc\u30af\uff09 \u3053\u306e\u30b3\u30fc\u30c9\u3092 Pythonic \u306b\u66f8\u304d\u76f4\u3057\u3066\u3044\u304d\u307e\u3059\u3002 \u3061\u306a\u307f\u306b\u3001 Python \u306f 3.3 \u3092\u4f7f\u3044\u307e\u3059\u3002 \u307e\u305a\u3001\u6700\u521d\u306e\u30b3\u30fc\u30c9 n = 4000 a = [] count = 1 for x in range(n): a.append([]) for y in range(n): a[x].append(count) count += 1 list(map(lambda e: e.reverse(), a)) print(a[-1][-1]) time \u30b3\u30de\u30f3\u30c9\u3067\u5b9f\u884c\u901f\u5ea6\u6e2c\u3063\u305f\u2026","provider_url":"https://hatena.blog","provider_name":"Hatena Blog","blog_url":"https://methane.hatenablog.jp/","url":"https://methane.hatenablog.jp/entry/2012/11/15/180948","html":"<iframe src=\"https://hatenablog-parts.com/embed?url=https%3A%2F%2Fmethane.hatenablog.jp%2Fentry%2F2012%2F11%2F15%2F180948\" title=\"\u52dd\u624b\u306b\u6dfb\u524a (C vs Python vs Ruby vs Haskell\uff08\u7121\u610f\u5473\u306a\u51e6\u7406de\u30d9\u30f3\u30c1\u30de\u30fc\u30af\uff09) - methane\u306e\u30d6\u30ed\u30b0\" class=\"embed-card embed-blogcard\" scrolling=\"no\" frameborder=\"0\" style=\"display: block; width: 100%; height: 190px; max-width: 500px; margin: 10px 0px;\"></iframe>","published":"2012-11-15 18:09:48","image_url":null,"version":"1.0","blog_title":"methane\u306e\u30d6\u30ed\u30b0","width":"100%","author_url":"https://blog.hatena.ne.jp/methane/","author_name":"methane"}