我是R新手,這是一個非常簡單的問題,我似乎無法解決... 所以我想製作一個包含A1〜A12,B1〜 B12,C1〜C12,D1〜D12,E1〜E12,F1〜F12,G1〜G12,H1〜H12。類似下面...製作r中的字符列表
[1] "A1" "A2" "A3" "A4" "A5" "A6" "A7" "A8" "A9" "A10" "A11" "A12" "B1" "B2" "B3" "B4" "B5" "B6" "B7" "B8"
[21] "B9" "B10" "B11" "B12" "C1" "C2" "C3" "C4" "C5" "C6" "C7" "C8" "C9" "C10" "C11" "C12" "D1" "D2" "D3" "D4"
[41] "D5" "D6" "D7" "D8" "D9" "D10" "D11" "D12" "E1" "E2" "E3" "E4" "E5" "E6" "E7" "E8" "E9" "E10" "E11" "E12"
[61] "F1" "F2" "F3" "F4" "F5" "F6" "F7" "F8" "F9" "F10" "F11" "F12" "G1" "G2" "G3" "G4" "G5" "G6" "G7" "G8"
[81] "G9" "G10" "G11" "G12" "H1" "H2" "H3" "H4" "H5" "H6" "H7" "H8" "H9" "H10" "H11" "H12"
我試圖使用Rep ...或創建LETTERS的矢量[1:8]和c(1:12)的一個單獨的載體,並試圖它們結合在一起...但我並不是很成功。
提前致謝!
其他問題有點關係...
所以我做這個之後,我想這個比較另一個列表。另一個列表可能看起來像這樣:
[1] "A1" "A2" "A3" "A5" "A6" "A7" "A8" "A9" "A10" "A11" "A12" "B1" "B2" "B3" "B4" "B5" "B6" "B7" "B8" "B9"
[21] "B10" "B11" "B12" "C1" "C2" "C3" "C4" "C5" "C6" "C7" "C8" "C9" "C10" "C11" "C12" "D1" "D2" "D3" "D4" "D5"
[41] "D6" "D7" "D8" "D9" "D10" "D11" "D12" "E1" "E2" "E3" "E4" "E5" "E6" "E7" "E8" "E9" "E10" "E11" "E12" "F1"
[61] "F2" "F3" "F4" "F5" "F6" "F7" "F8" "F9" "F10" "F11" "F12" "G1" "G2" "G3" "G4" "G5" "G6" "G7" "G8" "G9"
[81] "G10" "G11" "G12" "H1" "H2" "H3" "H4" "H5" "H6" "H7" "H8" "H9" "H10" "H11" "H12"
這不是很清楚,但此列表缺少「A4」。 通過比較這一個和我創建的具有所有96個元素的人之間的關係,我想知道哪個元素丟失了。我嘗試過使用像intersect和setdiffer這樣的函數,但是它們不會比較列表中的元素。
'setdiff(biglist,smallerlist)'你想要做什麼。在與原文無關的其他問題中編輯也不是一種好的形式。在這種情況下,我不會推薦一個新的問題,因爲它幾乎肯定是重複的:請參閱http://stackoverflow.com/questions/14440585/very-simple-subset-selection-in-r – thelatemail