{"id":28342,"date":"2022-12-30T07:50:33","date_gmt":"2022-12-30T02:20:33","guid":{"rendered":"https:\/\/jassweb.com\/solved\/solved-print-elements-of-list-horizontally-closed\/"},"modified":"2022-12-30T07:50:33","modified_gmt":"2022-12-30T02:20:33","slug":"solved-print-elements-of-list-horizontally-closed","status":"publish","type":"post","link":"https:\/\/jassweb.com\/solved\/solved-print-elements-of-list-horizontally-closed\/","title":{"rendered":"[Solved] Print elements of list horizontally [closed]"},"content":{"rendered":"

[ad_1]
\n<\/p>\n

\n
\n
<\/div>\n
\n
\n

Here’s a silly one-liner version:<\/p>\n

def print_table(seq, width=3):\n    print('\\n'.join([''.join(\n        [(str(u[-i]) if len(u) >= i else '').rjust(width) for u in seq])\n            for i in range(max(len(u) for u in seq), 0, -1)]))\n<\/code><\/pre>\n
And here’s the same algorithm in a somewhat more readable form:<\/p>\n
def get_cell(u, i):\n    return str(u[-i]) if len(u) >= i else ''\n\ndef print_table(seq, width=3):\n    maxlen = max(len(u) for u in seq)\n    for i in range(maxlen, 0, -1):\n        row = [get_cell(u, i).rjust(width) for u in seq]\n        print(''.join(row))\n\n#Test\nlista=[[2], [1, 2], [2], [3], [3]]\nlistb=[[4], [2, 5], [1, 3, 6], [7], [8]]\n\nprint(lista)\nprint_table(lista)\nprint('-' * 20)\n\nprint(listb)\nprint_table(listb)\n<\/code><\/pre>\n
output<\/strong><\/p>\n
[[2], [1, 2], [2], [3], [3]]\n     1         \n  2  2  2  3  3\n--------------------\n[[4], [2, 5], [1, 3, 6], [7], [8]]\n        1      \n     2  3      \n  4  5  6  7  8\n<\/code><\/pre>\n
I won’t explain exactly how this code works, since I suspect that this is a homework question.<\/p>\n<\/p><\/div>\n
<\/div>\n<\/div>\n
            1<\/span> <\/p><\/div>\n<\/div>\n
[ad_2]<\/p>\n
solved Print elements of list horizontally [closed] <\/p>\n","protected":false},"excerpt":{"rendered":"
[ad_1] Here’s a silly one-liner version: def print_table(seq, width=3): print(‘\\n’.join([”.join( [(str(u[-i]) if len(u) >= i else ”).rjust(width) for u in seq]) for i in range(max(len(u) for u in seq), 0, -1)])) And here’s the same algorithm in a somewhat more readable form: def get_cell(u, i): return str(u[-i]) if len(u) >= i else ” def print_table(seq, … 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":[349,482],"class_list":["post-28342","post","type-post","status-publish","format-standard","hentry","category-solved","tag-python","tag-python-3-x"],"yoast_head":"\n[Solved] Print elements of list horizontally [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-print-elements-of-list-horizontally-closed\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"[Solved] Print elements of list horizontally [closed] - JassWeb\" \/>\n<meta property=\"og:description\" content=\"[ad_1] Here’s a silly one-liner version: def print_table(seq, width=3): print('n'.join([''.join( [(str(u[-i]) if len(u) >= i else '').rjust(width) for u in seq]) for i in range(max(len(u) for u in seq), 0, -1)])) And here’s the same algorithm in a somewhat more readable form: def get_cell(u, i): return str(u[-i]) if len(u) >= i else '' def print_table(seq, ... 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