Question

我有一个data.table，我需要为同一个cID添加标签== 1和标签== 2之间的比率的其他列。我有能够做到这一点的代码，但结果是根据唯一“l”的数量减少的形式。但我需要一个包含重复项的完整列表。有什么建议？提前谢谢！

x   y   l   cID
0.03588851  0.081635056 1   1
0.952514891 0.82677373  1   1
0.722920691 0.687278396 1   1
0.772207687 0.743329599 2   1
0.682710551 0.946685728 1   2
0.795816439 0.024320077 2   2
0.50788885  0.106910923 2   2
0.145871035 0.802771467 2   2
0.092942384 0.335054397 1   3
0.439765866 0.199329139 1   4

重现

x = c(0.03588851,0.952514891,0.722920691,0.772207687,0.682710551,0.795816439,0.50788885,0.145871035,0.092942384,0.439765866)
y = c(0.081635056,0.82677373,0.687278396,0.743329599,0.946685728,0.024320077,0.106910923,0.802771467,0.335054397,0.199329139)
l = c(1,1,1,2,1,2,2,2,1,1)
cID = c(1,1,1,1,2,2,2,2,3,4)
dt <- data.table(x,y,l,cID)
dt[,sum(l == 1)/sum(l == 2), by = cID]

我需要获得看起来像这样的比率列

x   y   l   cID ratio
0.03588851  0.081635056 1   1   3
0.952514891 0.82677373  1   1   3
0.722920691 0.687278396 1   1   3
0.772207687 0.743329599 2   1   3
0.682710551 0.946685728 1   2   0.333333333
0.795816439 0.024320077 2   2   0.333333333
0.50788885  0.106910923 2   2   0.333333333
0.145871035 0.802771467 2   2   0.333333333
0.092942384 0.335054397 1   3   Inf
0.439765866 0.199329139 1   4   Inf

Answer 1

你非常接近。试试这个：

dt[, ratio := sum(l == 1) / sum(l == 2), by = cID]

计算值与data.table中特定标签的比率

1 个答案: