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作者 physheepy 在 PTT [ Python ] 看板的留言(推文), 共22則
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7F推: 很常用 你不常用可能是還在用c或其他語言思考?06/21 22:38
12F→: 你應該是在用python3 看的範例卻是python2的04/12 12:37
13F→: zip在python2 是產生new list of turples 所以我用04/12 12:37
14F→: python2可正常執行 python3中 zip改成 generator04/12 12:38
15F推: 你原本的code 加一行 zipobj2 = [ x for x in zipobj]04/12 12:40
16F→: 應該就可以了(我不確定 因為我是用python2 測試看看吧)04/12 12:40
17F→: 另外處理表格建議使用pandas04/12 12:41
18F→: 補充一下 雖然加一行可以解決你的問題 但這種用法是違04/12 12:49
19F→: 背generator精神的 (剛發現更簡單的寫法04/12 12:50
20F→: zipobj = list(zip(.....))04/12 12:51
5F推: 改樓上的 (input % 5 > 0) ? 5*((input/5)+1) : input03/31 23:02
8F→: 啊~~乾尬了 我發現我寫的是c++ XD03/31 23:04
12F推: a = 5*((input/5)+1) if input%5 >0 else input03/31 23:08
5F→: goo.gl/zLodFr03/28 20:19
1F→: 喔對了 其實用不到numpy 用list就可以了03/27 14:20
1F推: numpy應該可以吧 (看不太懂你到底要做什麼)03/26 18:25
5F→: 如樓上說的 或者讓 QPD_select回傳 report_fileselect03/07 12:40
6F→: add "return report_fileselect" in QPD_select03/07 12:41
1F→: 資料量不大的話 if df[i].loc[0,0] == "XXXX": print i02/19 02:42
6F→: 這樣不會影響結果吧? 是數字的欄位就不是你要找的02/19 11:17
7F→: warning 不是 error02/19 11:18
8F推: 那個warning只是提醒你 你想比對的欄位不是 unicode02/19 11:24
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