我没有找到任何指定如何使用时间列，id列和具有多个变量的列重新整形数据框的内容，我想在不同的列中使用这些变量。
如果只需要两个类别，那就很简单：
How to reshape data from long to wide format?
但是，我有：
geo time indic_na value
AT 2014Q1 B11 2556
BE 2014Q1 B11 1506.0
... ... ... ...
AT 2014Q1 B1G 72065.1
我希望：
geo time B11 B1G ...
AT 2014Q1 2556 72065.1 ...
AT 2013Q4 2535.4 ...
... ... ... ... ...
BE 2014Q1 1506.0 86513.0 ...
所以我希望indic_na中的每个唯一字符串都成为一个列变量。获取数据：
install.packages("SmarterPoland")
library(zoo)
library(SmarterPoland)
GDP <- getEurostatRCV(kod = "namq_gdp_c")
GDP$time = as.yearqtr(GDP$time)
GDP <- subset(GDP, (s_adj == "SWDA") & (unit == "MIO_EUR") & (time > "1989Q4"))
然后我尝试了：
testvector <- as.vector(unique(GDP$indic_na))
test <- reshape(data = GDP, direction = "long", idvar = "geo", timevar = "time", varying = testvector)
在maaany其他事情中＆＃34;变化＆＃34; ;-)我收到此错误消息：
猜测错误（变化）：
无法从名字中猜出时变变量
我感觉如此接近！但不知怎的，我不能告诉R变量在我的数据框的第3列。我在网上找到的所有例子在不同的列中都有不同的变量，或者只有id或时间和一列变量。
任何帮助都会很棒！
> dput(head(GDP))
structure(list(geo = structure(c(1L, 3L, 4L, 5L, 6L, 7L), .Names = c("SWDA,MIO_EUR,B11,AT",
"SWDA,MIO_EUR,B11,BE", "SWDA,MIO_EUR,B11,BG", "SWDA,MIO_EUR,B11,CH",
"SWDA,MIO_EUR,B11,CY", "SWDA,MIO_EUR,B11,CZ"), .Label = c("AT",
"BA", "BE", "BG", "CH", "CY", "CZ", "DE", "DK", "EA", "EA12",
"EA17", "EA18", "EE", "EL", "ES", "EU15", "EU27", "EU28", "FI",
"FR", "HR", "HU", "IE", "IS", "IT", "JP", "LT", "LU", "LV", "ME",
"MK", "MT", "NL", "NO", "PL", "PT", "RO", "RS", "SE", "SI", "SK",
"TR", "UK", "US"), class = "factor"), time = structure(c(2014,
2014, 2014, 2014, 2014, 2014), class = "yearqtr"), indic_na = structure(c(1L,
1L, 1L, 1L, 1L, 1L), .Names = c("SWDA,MIO_EUR,B11,AT", "SWDA,MIO_EUR,B11,BE",
"SWDA,MIO_EUR,B11,BG", "SWDA,MIO_EUR,B11,CH", "SWDA,MIO_EUR,B11,CY",
"SWDA,MIO_EUR,B11,CZ"), .Label = c("B11", "B111", "B112", "B1G",
"B1GM", "B1GM_XE", "B1GM_XI", "B1GM_XO", "B2G_B3G", "D1", "D2_M_D3",
"D21_M_D31", "P3", "P3_P5", "P3_S13", "P31_S13", "P31_S14", "P31_S14_S15",
"P31_S15", "P32_S13", "P5", "P51", "P52", "P52_P53", "P53", "P6",
"P7"), class = "factor"), value = c(2556.8, 1506, NA, NA, NA,
3056.1)), .Names = c("geo", "time", "indic_na", "value"), row.names = 7753:7758, class = "data.frame")
答案 0 :(得分：3)
感谢您提出这样一个明确的问题！对于新用户来说很少见。我建议reshape2超过reshape。
GDP <- subset(GDP, (s_adj == "SWDA") & (unit == "MIO_EUR") & (time > "1989Q4"),
select = c("geo", "time", "indic_na", "value"))
# Making your data match your example
library(reshape2)
GDP_wide <- dcast(GDP, geo + time ~ indic_na, value.var = "value")
> head(GDP_wide)
geo time B11 B111 B112 ...
1 AT 1990 Q1 -64.3 -1407.1 1337.6
2 AT 1990 Q2 -37.2 -1432.0 1450.3
3 AT 1990 Q3 -39.4 -1457.4 1544.2
4 AT 1990 Q4 -78.7 -1546.7 1592.7
5 AT 1991 Q1 -140.2 -1771.9 1583.0
6 AT 1991 Q2 -183.7 -1938.5 1568.3
答案 1 :(得分：2)
您也可以尝试：
library(dplyr)
library(tidyr)
GDP %>%
filter(s_adj=="SWDA" & unit=="MIO_EUR" & time >"1989Q4") %>%
select(geo, time, indic_na, value) %>%
spread(indic_na, value) %>%
mutate(time=as.yearqtr(time)) %>%
head(2)
geo time B11 B111 B112 B1G B1GM B1GM_XE B1GM_XI B1GM_XO
1 AT 1990 Q1 -64.3 -1407.1 1337.6 28198.6 31783.3 -132.0 0 0
2 AT 1990 Q2 -37.2 -1432.0 1450.3 28611.4 32215.3 -256.7 0 0
B2G_B3G D1 D21_M_D31 D2_M_D3 P3 P31_S13 P31_S14 P31_S14_S15 P31_S15
1 11123.2 16710.8 3592.1 3913.5 23858.2 3342.0 17493.5 17925.6 429.9
2 11304.7 16950.7 3613.1 3941.5 24150.3 3380.1 17696.8 18137.3 438.2
P32_S13 P3_P5 P3_S13 P5 P51 P52 P52_P53 P53 P6 P7
1 2614.0 31963.5 5969.7 8105.3 7544.3 440.6 516.1 75.5 11640.6 11785.3
2 2632.1 32329.2 6020.1 8178.9 7652.9 394.7 472.2 77.5 11916.3 11851.2
答案 2 :(得分：1)
作为参考，这里是基础R reshape方法：
## Subsetting as per @Gregor's answer
GDP <- subset(GDP, (s_adj == "SWDA") & (unit == "MIO_EUR") & (time > "1989Q4"),
select = c("geo", "time", "indic_na", "value"))
## The actual reshaping step
gdp_wide <- reshape(GDP, direction = "wide",
idvar=c("geo", "time"),
timevar="indic_na")
head(gdp_wide, 3)
# geo time value.B11 value.B111 value.B112 value.B1G value.B1GM
# 7753 AT 2014 Q1 2556.8 -352.6 4295.2 72065.1 79361.4
# 7754 BE 2014 Q1 1506.0 NA NA 86513.0 96789.0
# 7755 BG 2014 Q1 NA NA NA 8496.6 9881.7
# value.B1GM_XE value.B1GM_XI value.B1GM_XO value.B2G_B3G value.D1
# 7753 177 NA NA 30409.6 40402.3
# 7754 NA NA NA 36251.0 50573.0
# 7755 NA NA NA 4475.7 4133.9
# value.D21_M_D31 value.D2_M_D3 value.P3 value.P31_S13 value.P31_S14
# 7753 7351.4 8540.1 58941.2 9043.1 42602.2
# 7754 10276.0 9965.0 75733.0 NA 50434.0
# 7755 1370.1 1247.5 8099.9 879.7 6296.1
# value.P31_S14_S15 value.P31_S15 value.P32_S13 value.P3_P5 value.P3_S13
# 7753 43838.4 1221.8 6153.4 75305.2 15154.2
# 7754 51499.0 1065.0 NA 95283.0 24234.0
# 7755 6338.8 40.7 895.6 10230.8 1790.3
# value.P5 value.P51 value.P52 value.P52_P53 value.P53 value.P6 value.P7
# 7753 16364.0 16647.9 -281.3 68 349.3 46458.8 41903.8
# 7754 19550.0 19694.0 -144.0 -144 0.0 81501.0 79996.0
# 7755 2130.9 1975.9 72.0 72 NA 6850.2 7117.0
以下是您的方法，其中包含一些可以帮助您使用reshape的评论。但是，现在有dcast.data.table，reshape开始感觉很慢。
reshape(data = GDP, # This is correct
direction = "long", # Your data are already "long". You're going from
# "long" to "wide", so set it to "wide".
idvar = "geo", # idvars are like the row stubs in the resulting
# data. You can have multiple idvars (use `c()`)
timevar = "time", # timevars are like the colnames or how you want
# to spread the data out in a wide form.
varying = testvector) # If you only have one value column, this argument
# is rarely necessary. I generally set the
# "data" argument to just have the columns of
# interest so that I don't need to worry about
# this argument or to use the `drop` argument.