Question

我想为alpha，bravo和charlie创建一个滚动的2季度平均值（以及许多其他变量。研究带我到动物园并润滑包但似乎总是回到在一个变量或分组中滚动

set.seed(123)

dates <-  c("Q4'15", "Q1'16", "Q2'16","Q3'16", "Q4'16", "Q1'17", "Q2'17" ,"Q3'17", "Q4'17","Q1'18")

df <- data.frame(dates = sample(dates, 100,  replace = TRUE, prob=rep(c(.03,.07,.03,.08, .05),2)), 
                           alpha = rnorm(100, 5), bravo = rnorm(100, 10), charlie = rnorm(100, 15))

我正在寻找像

这样的东西

x <- df %>% mutate_if(is.numeric, funs(rollmean(., 2, align='right', fill=NA)))

期望的结果：整个＆＃34; Q4＆＃39; 15＆＃34;的加权平均值＆安培; ＆＃34; Q1＆＃39; 16＆＃34;，＆＃34; Q1＆＃39; 16＆＃34; ＆安培; ＆＃34; Q2＆＃39; 16＆＃34;等每列数据（alpha，bravo，charlie）。不寻找配对季度平均值的平均值。

以下是Q4＆＃39; 15＆amp;＆＃34; Q1＆＃39; 16＆＃34;的平均值。时间点

df %>% filter(dates %in% c("Q4'15", "Q1'16")) %>%  select(-dates) %>% summarise_all(mean)

Answer 1

我喜欢data.table，我有一个解决方案，但可能会更优雅。这就是我所拥有的：

数据

现在为data.table：

R> suppressMessages(library(data.table))
R> set.seed(123)
R> datesvec <- c("Q4'15", "Q1'16", "Q2'16","Q3'16", "Q4'16",
+               "Q1'17", "Q2'17" ,"Q3'17", "Q4'17","Q1'18")
R> df <- data.table(dates = sample(dates, 100,  replace = TRUE,
+                                 prob=rep(c(.03,.07,.03,.08, .05),2)),
+                  alpha = rnorm(100, 5),
+                  bravo = rnorm(100, 10),
+                  charlie = rnorm(100, 15))
R> df[ , ind := which(datesvec==dates), by=dates]
R> setkey(df, ind)  # optional but may as well
R> head(df)
   dates   alpha    bravo charlie ind
1: Q4'15 5.37964 11.05271 14.4789   1
2: Q4'15 7.05008 10.36896 15.0892   1
3: Q4'15 4.29080 12.12845 13.6047   1
4: Q4'15 5.00576  8.93667 13.3325   1
5: Q4'15 3.53936  9.81707 13.6360   1
6: Q1'16 3.45125 10.56299 16.0808   2
R>

这里的关键是我们需要恢复/维护您的数据表示所没有的宿舍的时间顺序。

按季度平均

使用data.table：

很容易

R> ndf <- df[ ,
+           .(qtr=head(dates,1),          # label of quarter
+             sa=sum(alpha),              # sum of a in quarter
+             sb=sum(bravo),              # sum of b in quarter
+             sc=sum(charlie),            # sum of c in quarter
+             n=.N),                      # number of observations
+           by=ind]
R> ndf
    ind   qtr      sa       sb       sc  n
 1:   1 Q4'15 25.2656  52.3039  70.1413  5
 2:   2 Q1'16 65.8562 132.6650 192.7921 13
 3:   3 Q2'16 10.3422  17.8061  31.3404  2
 4:   4 Q3'16 84.6664 168.1914 256.9010 17
 5:   5 Q4'16 41.3268  87.8253 139.5873  9
 6:   6 Q1'17 42.6196  85.4059 134.8205  9
 7:   7 Q2'17 76.5190 162.0784 241.2597 16
 8:   8 Q3'17 42.8254  83.2483 127.2600  8
 9:   9 Q4'17 68.1357 133.5794 198.1920 13
10:  10 Q1'18 37.0685  78.4107 120.2808  8
R>

暂停那些平均值

R> ndf[, `:=`(psa=shift(sa),               # previous sum of a
+            psb=shift(sb),               # previous sum of b
+            psc=shift(sc),                # previous sum of c
+            pn=shift(n))]                # previous nb of obs
R> ndf
    ind   qtr      sa       sb       sc  n     psa      psb      psc pn
 1:   1 Q4'15 25.2656  52.3039  70.1413  5      NA       NA       NA NA
 2:   2 Q1'16 65.8562 132.6650 192.7921 13 25.2656  52.3039  70.1413  5
 3:   3 Q2'16 10.3422  17.8061  31.3404  2 65.8562 132.6650 192.7921 13
 4:   4 Q3'16 84.6664 168.1914 256.9010 17 10.3422  17.8061  31.3404  2
 5:   5 Q4'16 41.3268  87.8253 139.5873  9 84.6664 168.1914 256.9010 17
 6:   6 Q1'17 42.6196  85.4059 134.8205  9 41.3268  87.8253 139.5873  9
 7:   7 Q2'17 76.5190 162.0784 241.2597 16 42.6196  85.4059 134.8205  9
 8:   8 Q3'17 42.8254  83.2483 127.2600  8 76.5190 162.0784 241.2597 16
 9:   9 Q4'17 68.1357 133.5794 198.1920 13 42.8254  83.2483 127.2600  8
10:  10 Q1'18 37.0685  78.4107 120.2808  8 68.1357 133.5794 198.1920 13
R>

当前和上一季度的平均值

R> ndf[is.finite(psa),                     # where we have valid data
+     `:=`(ra=(sa+psa)/(n+pn),            # total sum / total n == avg
+          rb=(sb+psb)/(n+pn),
+          rc=(sc+psc)/(n+pn))]
R> ndf[,c(1:2, 11:13)]
    ind   qtr      ra       rb      rc
 1:   1 Q4'15      NA       NA      NA
 2:   2 Q1'16 5.06233 10.27605 14.6074
 3:   3 Q2'16 5.07989 10.03141 14.9422
 4:   4 Q3'16 5.00045  9.78935 15.1706
 5:   5 Q4'16 4.84589  9.84680 15.2496
 6:   6 Q1'17 4.66369  9.62395 15.2449
 7:   7 Q2'17 4.76554  9.89937 15.0432
 8:   8 Q3'17 4.97268 10.22195 15.3550
 9:   9 Q4'17 5.28386 10.32513 15.4977
10:  10 Q1'18 5.00972 10.09476 15.1654
R>

利用两个季度的总和除以观察总数的事实与这两个季度的平均值相同的事实。（这反映了我之前想到的一个编辑）。

抽查

我们可以使用data.table的选择功能手动计算其中两行，我在这里选择索引为<1,2>和<4,5>的那些：

R> df[ ind <= 2, .(a=mean(alpha), b=mean(bravo), c=mean(charlie))]
         a      b       c
1: 5.06233 10.276 14.6074
R> df[ ind == 4 | ind == 5, .(a=mean(alpha), b=mean(bravo), c=mean(charlie))]
         a      b       c
1: 4.84589 9.8468 15.2496
R>

这很平常，由于data.table，该方法可以轻松扩展到数百万行。

PS：All in One

正如您提到的管道等，您可以使用链式data.table操作来编写所有这些内容。不是我喜欢的风格，但可能。以下内容创建了完全相同的内容，而没有像上面那样创建ndf临时：

## All in one
df[ , ind := which(datesvec==dates), by=dates][
   ,
    .(qtr=head(dates,1),          # label of quarter
      sa=sum(alpha),              # sum of a in quarter
      sb=sum(bravo),              # sum of b in quarter
      sc=sum(charlie),            # sum of c in quarter
      n=.N),                      # number of observations
    by=ind][
   ,
    `:=`(psa=shift(sa),               # previous sum of a
         psb=shift(sb),               # previous sum of b
         psc=shift(sc),                # previous sum of c
         pn=shift(n))][
    is.finite(psa),                     # where we have valid data
    `:=`(ra=(sa+psa)/(n+pn),            # total sum / total n == avg
         rb=(sb+psb)/(n+pn),
         rc=(sc+psc)/(n+pn))][
    ,c(1:2, 11:13)][]

滚动加权平均值跨越两个因子水平或时间点

1 个答案:

数据

按季度平均

暂停那些平均值

当前和上一季度的平均值

抽查

PS：All in One