{"id":21486,"date":"2022-11-13T21:15:51","date_gmt":"2022-11-13T15:45:51","guid":{"rendered":"https:\/\/jassweb.com\/solved\/solved-code-how-can-i-go-over-a-list-of-numbers-and-print-the-number-of-consecutive-increases-in-python-closed\/"},"modified":"2022-11-13T21:15:51","modified_gmt":"2022-11-13T15:45:51","slug":"solved-code-how-can-i-go-over-a-list-of-numbers-and-print-the-number-of-consecutive-increases-in-python-closed","status":"publish","type":"post","link":"https:\/\/jassweb.com\/solved\/solved-code-how-can-i-go-over-a-list-of-numbers-and-print-the-number-of-consecutive-increases-in-python-closed\/","title":{"rendered":"[Solved] Code: How can I go over a list of numbers and print the number of consecutive increases in Python? [closed]"},"content":{"rendered":"

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There are 4 things I’d like to mention. Here’s some code and its output:<\/p>\n

def srk_func(words):\n    current = []\n    lastc = []\n    for x in words:\n        if len(current) == 0: \n                current.append(int(x))\n        elif len(current) == 1:\n                if current[0] < int(x):\n                        current.append(int(x))\n                else:          \n                        if len(current) >= len(lastc):\n                                lastc = current\n                        current[:] = []\n                        current.append(int(x))\n        elif len(current) >= 2:\n                if current[-1] < int(x):\n                        current.append(int(x))\n                else:          \n                        if len(current) >= len(lastc):\n                                lastc = current\n                        elif len(current) < len(lastc):\n                                current[:] = []\n                        current[:] = []\n                        current.append(int(x))\n    return lastc\n\ndef jm_func(words):\n    current = []\n    lastc = []\n    for w in words:\n        x = int(w)\n        if not current:\n            # this happens only on the first element\n            current = [x]\n            continue\n        if x > current[-1]:\n            current.append(x)\n        else:\n            # no increase, so current is complete\n            if len(current) >= len(lastc):\n                lastc = current\n            current = [x]\n    # end of input, so current is complete\n    if len(current) >= len(lastc):\n        lastc = current\n    return lastc\n\ntests = \"\"\"\\\n    1\n    1 5\n    5 1\n    1 5 7\n    7 5 1\n    1 5 7 0\n    1 5 7 0 3\n    1 5 7 0 2 4 6 8\n    1 3 5 7 9 11 0 2\n    \"\"\"\n\nfor test in tests.splitlines():\n    wds = test.split()\n    print wds\n    print srk_func(wds)\n    print jm_func(wds)\n    print\n\n8<--------------------------------------------------\n\n['1']\n[]\n[1]\n\n['1', '5']\n[]\n[1, 5]\n\n['5', '1']\n[1]\n[1]\n\n['1', '5', '7']\n[]\n[1, 5, 7]\n\n['7', '5', '1']\n[1]\n[1]\n\n['1', '5', '7', '0']\n[0]\n[1, 5, 7]\n\n['1', '5', '7', '0', '3']\n[0, 3]\n[1, 5, 7]\n\n['1', '5', '7', '0', '2', '4', '6', '8']\n[0, 2, 4, 6, 8]\n[0, 2, 4, 6, 8]\n\n['1', '3', '5', '7', '9', '11', '0', '2']\n[0, 2]\n[1, 3, 5, 7, 9, 11]\n\n[]\n[]\n[]\n<\/code><\/pre>\n
Topic 1: Test your code.<\/p>\n
Topic 2: Redundancy: Your code for len(current) == 1<\/code> is functionally identical to your code for len >= 2, and the latter is bloated by having the following unnecessary two lines:<\/p>\n
  elif len(current) < len(lastc):\n        current[:] = []\n<\/code><\/pre>\n
You can combine the two cases; see my version.<\/p>\n
Topic 3: It often happens in this kind of algorithm where you are processing input and keeping some “state” (in this case, current and lastc) that you can’t immediately pack up and go home when you reach the end of the input; you need to do something with that state.<\/p>\n
Topic 4: This is going to get technical, but it’s a common trap for new Python players. <\/p>\n
>>> current = [1, 2, 3, 4, 5]\n>>> lastc = current # lastc and current refer to THE SAME LIST; no copying!\n>>> print current\n[1, 2, 3, 4, 5]\n>>> print lastc\n[1, 2, 3, 4, 5]\n>>> current[:] = [] # The list to which current refers is cleared\n>>> print current\n[]\n>>> print lastc # lastc refers to the same list\n[]\n<\/code><\/pre>\n
It’s better to just assign the name current<\/code> to a new list; see my code.<\/p>\n
Topic 5 (bonus extra): Consider using 4-space indentation, not 8. Causing both the horizontal and vertical scroll-bars to appear in an SO question or answer == FAIL \ud83d\ude42<\/p>\n<\/p><\/div>\n
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            1<\/span> <\/p><\/div>\n<\/div>\n
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solved Code: How can I go over a list of numbers and print the number of consecutive increases in Python? [closed] <\/p>\n","protected":false},"excerpt":{"rendered":"
[ad_1] There are 4 things I’d like to mention. Here’s some code and its output: def srk_func(words): current = [] lastc = [] for x in words: if len(current) == 0: current.append(int(x)) elif len(current) == 1: if current[0] < int(x): current.append(int(x)) else: if len(current) >= len(lastc): lastc = current current[:] = [] current.append(int(x)) elif len(current) … Read more<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[320],"tags":[540,738,349],"class_list":["post-21486","post","type-post","status-publish","format-standard","hentry","category-solved","tag-list","tag-numbers","tag-python"],"yoast_head":"\n[Solved] Code: How can I go over a list of numbers and print the number of consecutive increases in Python? [closed] - JassWeb<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/jassweb.com\/solved\/solved-code-how-can-i-go-over-a-list-of-numbers-and-print-the-number-of-consecutive-increases-in-python-closed\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"[Solved] Code: How can I go over a list of numbers and print the number of consecutive increases in Python? [closed] - JassWeb\" \/>\n<meta property=\"og:description\" content=\"[ad_1] There are 4 things I’d like to mention. Here’s some code and its output: def srk_func(words): current = [] lastc = [] for x in words: if len(current) == 0: current.append(int(x)) elif len(current) == 1: if current[0] < int(x): current.append(int(x)) else: if len(current) >= len(lastc): lastc = current current[:] = [] current.append(int(x)) elif len(current) ... 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