英文:
Conditionally returning a vector of some row values based on another column's row values
问题
我正在尝试在“lab”列的相应值为“A”时将“diff”列的行作为向量返回。但条件是，对于“BB”，不仅会删除相应的diff值，还会跳过上下行的两个即时值。
从上述示例中，预期的输出如下：
> res
[1] 0.0314439 -1.2708040 0.4045876 -0.1069435
请问还有其他需要帮助的地方吗？谢谢！
英文:
Here is a reproducible data set:
set.seed(55);
data <- rnorm(12);
dates <- as.POSIXct("2019-03-18 10:30:00", tz = "CET") + 0:(length(data)-1)*60;
R <- xts(x = data, order.by = dates) %>%
sample(size = 10) %>%
fortify.zoo()
colnames(R) <- c("Time", "Rf");
R$lab <- "A"
R$lab[c(5, length(R$lab))] <- "BB"
R$diff <- c(NA, diff(R$Rf))
Output looks like:
> R
Time Rf lab diff
1 2019-03-18 10:30:00 0.120139084 A NA
2 2019-03-18 10:32:00 0.151582984 A 0.0314439
3 2019-03-18 10:33:00 -1.119221005 A -1.2708040
4 2019-03-18 10:34:00 0.001908206 A 1.1211292
5 2019-03-18 10:36:00 -0.505343855 BB -0.5072521
6 2019-03-18 10:37:00 -0.099234393 A 0.4061095
7 2019-03-18 10:38:00 0.305353199 A 0.4045876
8 2019-03-18 10:39:00 0.198409703 A -0.1069435
9 2019-03-18 10:40:00 -0.048910950 A -0.2473207
10 2019-03-18 10:41:00 -0.843233767 BB -0.7943228
I am trying to return the rows of column "diff" as a vector when the corresponding value of "lab" column is "A". But the condition is, for "BB" not only the corresponding diff value is dropped but also the two immediate values from upper and lower rows are also skipped.
Of the above example, following output is expected:
> res
[1] 0.0314439 -1.2708040 0.4045876 -0.1069435
Can you kindly help? Thanks
答案1
得分: 2
你可以尝试
inds <- which(R$lab == "BB")
R$diff[-unique(c(inds - 1, inds, inds + 1))]
或者如@Rui Barradas提到的
R$diff[-sapply(inds, `+`, -1:1)]
英文:
You can try
inds <- which(R$lab == "BB")
R$diff[-unique(c(inds - 1, inds, inds + 1))]
Or as @Rui Barradas mentioned
R$diff[-sapply(inds, `+`, -1:1)]
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