这是我的数据框：
structure(list(Year = c(1975L, 1975L, 1975L, 1975L, 1975L, 1975L,
1975L, 1975L, 1975L, 1975L, 1975L, 1975L, 1976L, 1976L), Month = c(1L,
2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 1L, 2L), A = c(0.419727177,
0.411522634, 0.415627598, 0.425350915, 0.431778929, 0.455892409,
0.464252553, 0.473933649, 0.48947626, 0.481231954, 0.495049505,
0.49419323, 0.4927322, 0.493778392), log.S.grb = c(-0.86815035734372,
-0.887891256732457, -0.877965616764487, -0.854840768392932, -0.839841560146749,
-0.785498442482632, -0.767326579572197, -0.746687948097975, -0.714419316655485,
-0.731405892265036, -0.703097511313113, -0.704828684428783, -0.707789457396815,
-0.705668461635217), D = c(10.2641, 9.7704, 9.3694, 9.2403, 9.4459,
9.4826, 10.4272, 10.3805, 10.4835, 11.4103, 10.988, 10.7708,
9.2987, 8.6161), E = c(5.68, 5.4, 5.53, 5.5, 5.2, 5.86, 6.25,
6.36, 6.58, 5.51, 5.54, 5.2, 4.73, 5), J = c(1.15663289, 1.05923536,
0.938740721, 0.890710069, 1.012043355, 0.843618397, 0.850583558,
0.957856493, 0.888553262, 1.391339534, 1.309574432, 1.322714922,
1.247746749, 0.894350421), st3_st.grb = c(NA, NA, NA, 1.33095889507878,
4.80496965857083, 9.24671742818556, 8.75141888207357, 9.31536120487736,
7.10791258271466, 3.59206873071611, 4.35904367848617, 0.959063222670167,
2.36164348682203, -0.257095032210358), idiff_3m.grb = c(1.08206512614859,
1.03439401057093, 0.909556730440464, 0.886569121738501, 1.00676358283349,
0.856864819463321, 0.983497311932147, 0.946390153374566, 0.917819573656709,
1.38813518035337, 1.28343598984861, 1.31491300418167, 1.08566149047631,
0.860507535403032), EXCESS = c("NA", "NA", "NA", "NA", "NA",
"NA", "NA", "NA", "NA", "NA", "NA", "NA", "NA", "NA")), row.names = c(NA,
14L), class = "data.frame")
我想要的是通过使用
dplyr（也许是mutate函数）进行以下计算：在EXCESS列中将是(idiff_3m.grb - st3_st.grb)操作的值，但是idiff_3m.grb的第一个值将是第三个值，并且st3_st.grb的第一个值将是第一个。结果将开始在EXCESS列的第三个位置上填充，依此类推。换句话说，第一个操作是：
(1.0820651 - 1.3309589) = -0.2488938 = EXCESS[3,] This first result will be in the third position of EXCESS column.
最后，我将在EXCESS列中用
"NA"填充三个位置（前三个）。有什么帮助吗？
最佳答案
您可以使用lag函数移动另一列，然后执行计算：
library(dplyr)
df %>%
mutate(idiff_3m.grb_lag3 = lag(idiff_3m.grb, 3),
EXCESS = idiff_3m.grb_lag3 - st3_st.grb)
返回：
Year Month A log.S.grb D E J st3_st.grb idiff_3m.grb EXCESS idiff_3m.grb_lag3
1 1975 1 0.4197272 -0.8681504 10.2641 5.68 1.1566329 NA 1.0820651 NA NA
2 1975 2 0.4115226 -0.8878913 9.7704 5.40 1.0592354 NA 1.0343940 NA NA
3 1975 3 0.4156276 -0.8779656 9.3694 5.53 0.9387407 NA 0.9095567 NA NA
4 1975 4 0.4253509 -0.8548408 9.2403 5.50 0.8907101 1.3309589 0.8865691 -0.24889377 1.0820651
5 1975 5 0.4317789 -0.8398416 9.4459 5.20 1.0120434 4.8049697 1.0067636 -3.77057565 1.0343940
6 1975 6 0.4558924 -0.7854984 9.4826 5.86 0.8436184 9.2467174 0.8568648 -8.33716070 0.9095567
7 1975 7 0.4642526 -0.7673266 10.4272 6.25 0.8505836 8.7514189 0.9834973 -7.86484976 0.8865691
8 1975 8 0.4739336 -0.7466879 10.3805 6.36 0.9578565 9.3153612 0.9463902 -8.30859762 1.0067636
9 1975 9 0.4894763 -0.7144193 10.4835 6.58 0.8885533 7.1079126 0.9178196 -6.25104776 0.8568648
10 1975 10 0.4812320 -0.7314059 11.4103 5.51 1.3913395 3.5920687 1.3881352 -2.60857142 0.9834973
11 1975 11 0.4950495 -0.7030975 10.9880 5.54 1.3095744 4.3590437 1.2834360 -3.41265353 0.9463902
12 1975 12 0.4941932 -0.7048287 10.7708 5.20 1.3227149 0.9590632 1.3149130 -0.04124365 0.9178196
13 1976 1 0.4927322 -0.7077895 9.2987 4.73 1.2477467 2.3616435 1.0856615 -0.97350831 1.3881352
14 1976 2 0.4937784 -0.7056685 8.6161 5.00 0.8943504 -0.2570950 0.8605075 1.54053102 1.2834360
您可以根据需要删除滞后列，也可以使用
df %>% mutate(EXCESS = lag(idiff_3m.grb, 3) - st3_st.grb)。