使用多个可变变量和不平衡的时间从宽变长答案

【问题标题】：Reshaping wide to long with multiple varying variables and unbalaned time使用多个可变变量和不平衡的时间从宽变长
【发布时间】：2019-05-17 18:17:42
【问题描述】：

我有两组不同的变量：年百分比和年。年百分比从 1999 年开始到 2012 年结束，但年份从 1999 年开始到 2013 年。

countrylabel annualpercentageshare.1999 year1990 year1991 year1992
1      Austria                         NA       NA       NA       NA
2      Belgium                         NA       NA       NA       NA
3     Bulgaria                   48.20000       NA       NA       NA
4      Estonia                         NA       NA       NA       NA
5       France                   47.52853       NA       NA       NA
6      Germany                         NA       NA       NA       NA

类似的东西。

我已经尝试过这段代码：

merge_data2 <- reshape(merge_data2, varying = list(2:ncol(merge_data2)), 
                       v.names = c("percentageshare", "Year"),
                       idvar = "countrylabel", direction = "long", times = 1990:2013)

但我收到此错误消息：

“reshapeLong 中的错误（数据，idvar = idvar，timevar = timevar，varying = varying，： 'lengths(variing)' 必须全部匹配 'length(times)'"

编辑：我想要这样的数据框：

countrylabel    time      annualpercentageshare        year
Austria          1990            NA                      NA
Austria          1991            NA                      NA

【问题讨论】：

几乎就是reshape2::melt(merge_data2, id = "countrylabel")。如果您需要填写缺失的年份，那是第二步。
或者，dplyr: gather(merge_data2, -countrylabel, key = metric, value = Value)
@Mako212：它不起作用。它给了我三个变量：countrylabel、variable 和 value，“variable”只包含“annualpercentageshare.1999”直到“annualpercentageshare.2012”
@Rainroad 它还应该包含所有“年份”列的行。
@Mako212：对不起，我想我的问题不够清楚。我刚刚编辑了我的问题以显示我想要的数据框格式。

标签： r dataframe reshape

【解决方案1】：

library(tidyr); library(dplyr)
df %>%
  gather(variable, value, -countrylabel) %>%
  separate("variable", into = c("stat", "time"), sep = -4) %>%
  spread(stat, value)

输出

   countrylabel time annualpercentageshare. year
1       Austria 1990                     NA   NA
2       Austria 1991                     NA   NA
3       Austria 1992                     NA   NA
4       Austria 1999                     NA   NA
5       Belgium 1990                     NA   NA
6       Belgium 1991                     NA   NA
7       Belgium 1992                     NA   NA
8       Belgium 1999                     NA   NA
9      Bulgaria 1990                     NA   NA
10     Bulgaria 1991                     NA   NA
11     Bulgaria 1992                     NA   NA
12     Bulgaria 1999               48.20000   NA
13      Estonia 1990                     NA   NA
14      Estonia 1991                     NA   NA
15      Estonia 1992                     NA   NA
16      Estonia 1999                     NA   NA
17       France 1990                     NA   NA
18       France 1991                     NA   NA
19       France 1992                     NA   NA
20       France 1999               47.52853   NA
21      Germany 1990                     NA   NA
22      Germany 1991                     NA   NA
23      Germany 1992                     NA   NA
24      Germany 1999                     NA   NA

【讨论】：

【解决方案2】：

reshape 喜欢"."，所以首先我们在year* 变量中插入一个。

names(d) <- gsub("year", "year.", names(d))

现在，在我们给出了reshape 缺失的列和order 之后，

d$annualpercentage.2002 <- NA
d$year.1999 <- NA
d <- d[c(1, order(names(d)[-1]) + 1)]

您的想法通过在列表中定义 varying 中的不同列排序来实现：

res <- reshape(d, varying=list(2:5, 6:9), direction="long", idvar="countrylabel", 
               times=1999:2002, v.names=c("annualpercentage", "year"))
res
#                  countrylabel time annualpercentage        year
# Austria.1999          Austria 1999               NA          NA
# Belgium.1999          Belgium 1999               NA          NA
# Bulgaria.1999        Bulgaria 1999       -0.6806495          NA
# Estonia.1999          Estonia 1999               NA          NA
# France.1999            France 1999               NA          NA
# Germany.1999          Germany 1999               NA          NA
# Switzerland.1999  Switzerland 1999       -1.8497570          NA
# Austria.2000          Austria 2000       -0.6033900  0.14970015
# Belgium.2000          Belgium 2000               NA -0.49201756
# Bulgaria.2000        Bulgaria 2000        0.8263925 -0.36320990
# Estonia.2000          Estonia 2000               NA -2.51032544
# France.2000            France 2000               NA  0.57800624
# Germany.2000          Germany 2000               NA -0.52295712
# Switzerland.2000  Switzerland 2000        0.2783076  0.25616728
# Austria.2001          Austria 2001       -2.6962484 -0.15375642
# Belgium.2001          Belgium 2001        1.3088577  0.72528621
# Bulgaria.2001        Bulgaria 2001               NA          NA
# Estonia.2001          Estonia 2001               NA -0.05563662
# France.2001            France 2001        0.2224629  0.74205086
# Germany.2001          Germany 2001               NA -0.01185349
# Switzerland.2001  Switzerland 2001        0.8354322 -1.40826638
# Austria.2002          Austria 2002               NA          NA
# Belgium.2002          Belgium 2002               NA  1.60874778
# Bulgaria.2002        Bulgaria 2002               NA          NA
# Estonia.2002          Estonia 2002               NA  0.55866704
# France.2002            France 2002               NA -1.59866472
# Germany.2002          Germany 2002               NA -0.11217415
# Switzerland.2002  Switzerland 2002               NA          NA

数据

d <- structure(list(countrylabel = c("Austria", "Belgium", "Bulgaria", 
"Estonia", "France", "Germany", "Switzerland"), annualpercentage.1999 = c(NA, 
-2.58060150400384, -0.0623757258909573, 0.267776001395166, NA, 
NA, 0.048219924249952), annualpercentage.2000 = c(NA, -0.249416955035044, 
1.3525450891501, 1.04446768824697, NA, -0.0582347596434839, -0.891400228849837
), annualpercentage.2001 = c(1.82469277697851, NA, NA, 1.04231605324821, 
NA, -0.900145118946308, -1.19320727433597), year2000 = c(0.633712375393134, 
NA, 1.24760861316098, -0.092964787061478, -0.59403260962332, 
NA, -0.650348234181285), year2001 = c(0.587318286831079, NA, 
NA, 0.348890470222513, NA, NA, NA), year2002 = c(0.0645316087966406, 
-0.279456557428068, NA, NA, -0.0627400036074545, 1.30419117694731, 
-0.484654596062051)), row.names = c(NA, -7L), class = "data.frame")

【讨论】：