Question

在之前的一个问题（for loop in irregular time series）中，我提出了一个数据框（参见下面的dput），其中用户rnso给出了以下内容，这对我来说非常有用。

for(ss in unique(mydf$site_id)){
  for(cc in 3:12){
    # do whatever function
    print(max(mydf[mydf$site_id == ss, cc],na.rm=TRUE))
  } }

> [1] 304 [1] 16.8 [1] 8.43 [1] 286 [1] 2 [1] 36 [1] 93 [1] 30 [1] 5.98
> [1] 69 [1] -38 [1] 14.7 [1] 7.85 [1] 515 [1] 2 [1] 18 [1] 180 [1] 106
> [1] 0.1 [1] 655'

我一直在使用

idx <- max(seq(along=data)) 
lastx <- signif(data[idx], digits=3)

识别每个mydf [ss，cc]的最后一个数据点，效果很好。

问题：如何进一步子集mydf [idx]来为每个idx提取日期？我已经尝试了一系列排列，并且通常会出现一些“不正确的尺寸”错误。谢谢大家！

数据：

mydf <- structure(list(site_id = c("39ADA00070", "39ADA00070", "39ADA00070", 
"39ADA00070", "39ADA00070", "39ADA00070", "39ADA00070", "39ADA00070", 
"39ADA00070", "39ADA00070", "39ADA00070", "39ADA00070", "39ADA00070", 
"39ADA00070", "39ADA00070", "39ADA00070", "39ADA00070", "39ADA00070", 
"39ADA00070", "39ADA00070", "39ADA00070", "39ADA00070", "39ADA00070", 
"39ADA00070", "39ADA00070", "39ADA00070", "39ADA00070", "39ADA00070", 
"39ADA00070", "39ADA00070", "39ALL00184", "39ALL00184", "39ALL00184", 
"39ALL00184", "39ALL00184", "39ALL00184", "39ALL00184", "39ALL00184", 
"39ALL00184", "39ALL00184", "39ALL00184", "39ALL00184", "39ALL00184", 
"39ALL00184", "39ALL00184", "39ALL00184", "39ALL00184", "39ALL00184", 
"39ALL00184", "39ALL00184", "39ALL00184", "39ALL00184", "39ALL00184"
), date = structure(c(6339, 8594, 9293, 9441, 10014, 10604, 11080, 
11821, 12717, 12907, 13081, 13277, 13459, 13635, 13822, 14012, 
14207, 14207, 14355, 14564, 14704, 14917, 15105, 15271, 15478, 
15644, 15833, 15834, 16009, 16203, 7783, 8406, 8554, 8686, 9034, 
9260, 9632, 9777, 10002, 10491, 10491, 11060, 11585, 12145, 12145, 
12696, 13242, 13242, 13775, 14363, 14881, 15428, 15974), class = "Date"), 
    var1 = c(NA, NA, NA, NA, NA, NA, NA, NA, NA, 159L, 148L, 
    149L, 134L, 179L, 205L, 193L, 109L, 109L, 177L, 75L, 272L, 
    150L, 115L, 232L, 230L, 183L, 159L, 159L, 304L, 220L, NA, 
    NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, 
    -98L, -98L, -38L, -74L, -74L, -80L, -48L), var2 = c(NA, NA, 
    NA, NA, NA, NA, NA, NA, 16.8, 16.8, 14.5, 14.2, 15.1, 14.5, 
    15, 15.2, 13.2, 13.2, 15, 15.2, 15.1, 14.4, 14.8, 15.2, 16.3, 
    NA, 14.3, 14.3, 15.6, 14.8, NA, 12, 14.7, NA, 14.6, NA, 13.7, 
    12.3, 12.5, 13.5, 13.5, 12.5, 13.1, 14.2, 14.2, 14.1, 12.5, 
    12.5, 13.5, 12.7, 12.6, 12.5, 12.6), var3 = c(NA, NA, NA, 
    NA, NA, NA, NA, NA, 7.35, 7.85, 7.5, 7.47, 7.62, 7.08, 7.08, 
    7.2, 7.4, 7.4, 7.26, 7.05, 6.56, 7.2, 7.42, 6.5, 7.81, 8.43, 
    7.57, 7.57, 7.42, 7.72, NA, 6.58, 6.8, NA, 7.75, NA, 7.06, 
    6.77, 6.41, 6.84, 6.84, 7.85, 7.13, 7.26, 7.26, 7.06, 7.14, 
    7.14, 7.11, 6.9, 7.11, 7.2, 7.1), var4 = c(NA, 283L, 216L, 
    223L, 256L, 165L, 192L, 216L, 173L, 216L, 179L, 282L, 146L, 
    227L, 141L, 210L, 160L, 162L, 157L, 140L, 235L, 166L, 216L, 
    NA, 162L, 193L, 286L, 274L, 163L, 209L, NA, 304L, 321L, 293L, 
    398L, 302L, 301L, 282L, 288L, 292L, 292L, 302L, 515L, 309L, 
    309L, 323L, 338L, 295L, 280L, 279L, 325L, 328L, 322L), var5 = c(NA, 
    NA, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 
    2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 
    2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2), var6 = c(NA, NA, 
    29L, 32L, 36L, 24L, 25L, 29L, 27L, 27L, 24L, 32L, 21L, 27L, 
    21L, 26L, 23L, 24L, 25L, 20L, 24L, 22L, 28L, 24L, 20L, 23L, 
    30L, 29L, 21L, 24L, 15L, 15L, 18L, 15L, 15L, 15L, 15L, 15L, 
    15L, 15L, 15L, 15L, 15L, 15L, 15L, 15L, 15L, 15L, 15L, 15L, 
    15L, 15L, 15L), var7 = c(NA, NA, 77, 83, 87, 66, 73, 73, 
    65, 76, 69, 93, 60, 76, 56, 77, 67, 68, 68, 60, 67, 63, 82, 
    69, 56, 68, 85, 83, 59, 68.2, 157, 159, 164, 169, 155, 176, 
    156, 156, 162, 162, 162, 160, 180, 163, 163, 158, 168, 171, 
    162, 167, 177, 167, 168), var8 = c(NA, NA, 25, 26, 29, 21, 
    22, 23, 20, 23, 21, 30, 17, 24, 16, 23, 20, 20, 21, 17, 23, 
    18, 25, 20, 17, 21, 27, 27, 17, 20.9, 91, 89, 96, 92, 86, 
    100, 89, 91, 92, 94, 94, 91, 97, 91, 91, 92, 98, 99, 94, 
    100, 106, 98, 100), var9 = c(1.02, 1, 0.37, 0.48, 0.88, 0.16, 
    0.17, 0.24, 0.25, 5.98, 0.26, 0.54, 0, 0.19, 0, 0.18, 0.14, 
    0.13, 0.16, 0.11, 0.19, 0.16, 0.26, NA, 0.11, 0.27, 0.19, 
    0.19, 0.1, 0.1, 0.1, 0.1, 0.1, 0.1, 0.1, NA, 0.1, 0.1, 0.1, 
    0.1, 0.1, 0.1, 0.1, 0.1, 0.1, 0.1, 0.1, 0.1, 0, 0, 0.1, 0.1, 
    0.1), var10 = c(50, 48, 64, 55, 52, 64, 69, 63.3, 56.1, 40.6, 
    58.6, 43.9, 62.2, 51.9, 55.6, 53.4, 61.3, 61, 61.1, 61.9, 
    51.5, 60.7, 52.2, NA, 66, 52.8, 46.8, 47.5, 59.2, 53.4, NA, 
    560, 650, 540, 548, 655, 565, 531, 540, 501, 501, 531, 535, 
    547, 547, 492, 537, 542, 512, 542, 548, 581, 540)), class = "data.frame", row.names = c(NA, 
-53L), .Names = c("site_id", "date", "var1", "var2", "var3", 
"var4", "var5", "var6", "var7", "var8", "var9", "var10"))

Answer 1

我将如何做到这一点：

library(data.table)
library(reshape2)
dcast(melt(setDT(mydf),id.vars='site_id')
      [,max(value,na.rm=TRUE),'site_id,variable'],
      site_id ~variable)

     site_id  date var1 var2 var3 var4 var5 var6 var7 var8 var9 var10
1 39ADA00070 16203  304 16.8 8.43  286    2   36   93   30 5.98    69
2 39ALL00184 15974  -38 14.7 7.85  515    2   18  180  106 0.10   655

我正在使用data.table进行分组操作
将数据放在长格式中，因为您要在多列中执行相同的操作
为每个变量（偶数日期）获取最大值
使用dcast将您的数据再次重新整理为宽屏格式。

不规则时间序列：最后一个数据点串联

数据：

1 个答案: