Question

我有两个动物园对象 Z1

                T1      T2      T3
2013-01-18      20      15      21
2013-01-20      30      18      17
2013-01-21      10      21      24
2013-01-22      15      17      15
2013-01-23      18      16      18

Z2

                T1          T2          T3
2013-01-18      0.05        0.01        0.09
2013-01-20      0.02        0.05        0.06
2013-01-21      0.06        0.02        0.07
2013-01-22      0.07        0.06        0.08
2013-01-23      0.01        0.06        0.05

我想把输出作为

                T1                 T2                   T3
2013-01-18      myfunc(20,0.05)     myfunc(15,0.01)     myfunc(21,0.09)
2013-01-20      myfunc(30,0.02)     myfunc(18,0.05)     myfunc(17,0.06)
2013-01-21      myfunc(10,0.06)     myfunc(21,0.02)     myfunc(24,0.07)
2013-01-22      myfunc(15,0.07)     myfunc(17,0.06)     myfunc(15,0.08)
2013-01-23      myfunc(18,0.01)     myfunc(16,0.06)     myfunc(18,0.05)

似乎使用apply / rollapply会完成任务，但由于我是R的新手，我无法做到这一点。

更新
假设我也有数据框

Name Quantity Base
 T1   100      T3
 T2   150      T3
 T3   160      T3

输出需要像这样

                T1                     T2                   T3
2013-01-18      myfunc(20,0.05,100)     myfunc(20,0.01,150)     myfunc(20,0.09,160)
2013-01-20      myfunc(30,0.02,100)     myfunc(30,0.05,150)     myfunc(30,0.06,160)
2013-01-21      myfunc(10,0.06,100)     myfunc(10,0.02,150)     myfunc(10,0.07,160)
2013-01-22      myfunc(15,0.07,100)     myfunc(15,0.06,150)     myfunc(15,0.08,160)
2013-01-23      myfunc(18,0.01,100)     myfunc(18,0.06,150)     myfunc(18,0.05,160)

请注意，现在第一个参数也基于数据帧的基本列

Answer 1

你可以尝试

z3 <- z1
z3[] <- mapply(function(x,y,i) myfunc(x[,i], y[,i]), 
                 list(z1), list(z2), seq_len(ncol(z1)))

更新

如果只有一个data.frame（“df1”）并且需要将其传递给myfunc

z3[] <- mapply(function(x,y,z, i) myfunc(x[,i], y[,i], z),
     list(z1), list(z2), split(df1$quantity, df1$name), 
                                  seq_len(ncol(z1)))

操作两个时间序列数据

1 个答案:

更新