[英]Reshaping data.frame from wide to long format
我在將data.frame
從寬表轉換為長表時遇到了一些麻煩。 目前它看起來像這樣:
Code Country 1950 1951 1952 1953 1954
AFG Afghanistan 20,249 21,352 22,532 23,557 24,555
ALB Albania 8,097 8,986 10,058 11,123 12,246
現在我想把這個data.frame
變成一個長的data.frame
。 是這樣的:
Code Country Year Value
AFG Afghanistan 1950 20,249
AFG Afghanistan 1951 21,352
AFG Afghanistan 1952 22,532
AFG Afghanistan 1953 23,557
AFG Afghanistan 1954 24,555
ALB Albania 1950 8,097
ALB Albania 1951 8,986
ALB Albania 1952 10,058
ALB Albania 1953 11,123
ALB Albania 1954 12,246
正如某些人在類似問題中所建議的那樣,我已經查看並嘗試使用melt()
和reshape()
函數。 但是,到目前為止,我只得到混亂的結果。
如果可能的話,我想用reshape()
function 來做,因為它看起來更容易處理。
三種替代解決方案:
1)使用data.table :
您可以使用與reshape2
包中相同的melt
功能(這是一個擴展和改進的實現)。 來自data.table
的melt
還具有比來自reshape2
的melt
-function 更多的參數。 例如,您還可以指定變量列的名稱:
library(data.table)
long <- melt(setDT(wide), id.vars = c("Code","Country"), variable.name = "year")
這使:
> long Code Country year value 1: AFG Afghanistan 1950 20,249 2: ALB Albania 1950 8,097 3: AFG Afghanistan 1951 21,352 4: ALB Albania 1951 8,986 5: AFG Afghanistan 1952 22,532 6: ALB Albania 1952 10,058 7: AFG Afghanistan 1953 23,557 8: ALB Albania 1953 11,123 9: AFG Afghanistan 1954 24,555 10: ALB Albania 1954 12,246
一些替代符號:
melt(setDT(wide), id.vars = 1:2, variable.name = "year")
melt(setDT(wide), measure.vars = 3:7, variable.name = "year")
melt(setDT(wide), measure.vars = as.character(1950:1954), variable.name = "year")
2)使用tidyr :
library(tidyr)
long <- wide %>% gather(year, value, -c(Code, Country))
一些替代符號:
wide %>% gather(year, value, -Code, -Country)
wide %>% gather(year, value, -1:-2)
wide %>% gather(year, value, -(1:2))
wide %>% gather(year, value, -1, -2)
wide %>% gather(year, value, 3:7)
wide %>% gather(year, value, `1950`:`1954`)
3) 使用reshape2 :
library(reshape2)
long <- melt(wide, id.vars = c("Code", "Country"))
給出相同結果的一些替代符號:
# you can also define the id-variables by column number
melt(wide, id.vars = 1:2)
# as an alternative you can also specify the measure-variables
# all other variables will then be used as id-variables
melt(wide, measure.vars = 3:7)
melt(wide, measure.vars = as.character(1950:1954))
筆記:
數據的另一個問題是這些值將被 R 作為字符值讀取(作為數字中的,
的結果)。 您可以使用gsub
和as.numeric
修復它:
long$value <- as.numeric(gsub(",", "", long$value))
或直接使用data.table
或dplyr
:
# data.table
long <- melt(setDT(wide),
id.vars = c("Code","Country"),
variable.name = "year")[, value := as.numeric(gsub(",", "", value))]
# tidyr and dplyr
long <- wide %>% gather(year, value, -c(Code,Country)) %>%
mutate(value = as.numeric(gsub(",", "", value)))
數據:
wide <- read.table(text="Code Country 1950 1951 1952 1953 1954
AFG Afghanistan 20,249 21,352 22,532 23,557 24,555
ALB Albania 8,097 8,986 10,058 11,123 12,246", header=TRUE, check.names=FALSE)
reshape()
需要一段時間來適應,就像melt
/ cast
一樣。 這是一個重塑的解決方案,假設您的數據框稱為d
:
reshape(d,
direction = "long",
varying = list(names(d)[3:7]),
v.names = "Value",
idvar = c("Code", "Country"),
timevar = "Year",
times = 1950:1954)
使用tidyr_1.0.0
,另一個選項是pivot_longer
library(tidyr)
pivot_longer(df1, -c(Code, Country), values_to = "Value", names_to = "Year")
# A tibble: 10 x 4
# Code Country Year Value
# <fct> <fct> <chr> <fct>
# 1 AFG Afghanistan 1950 20,249
# 2 AFG Afghanistan 1951 21,352
# 3 AFG Afghanistan 1952 22,532
# 4 AFG Afghanistan 1953 23,557
# 5 AFG Afghanistan 1954 24,555
# 6 ALB Albania 1950 8,097
# 7 ALB Albania 1951 8,986
# 8 ALB Albania 1952 10,058
# 9 ALB Albania 1953 11,123
#10 ALB Albania 1954 12,246
df1 <- structure(list(Code = structure(1:2, .Label = c("AFG", "ALB"), class = "factor"),
Country = structure(1:2, .Label = c("Afghanistan", "Albania"
), class = "factor"), `1950` = structure(1:2, .Label = c("20,249",
"8,097"), class = "factor"), `1951` = structure(1:2, .Label = c("21,352",
"8,986"), class = "factor"), `1952` = structure(2:1, .Label = c("10,058",
"22,532"), class = "factor"), `1953` = structure(2:1, .Label = c("11,123",
"23,557"), class = "factor"), `1954` = structure(2:1, .Label = c("12,246",
"24,555"), class = "factor")), class = "data.frame", row.names = c(NA,
-2L))
使用重塑包:
#data
x <- read.table(textConnection(
"Code Country 1950 1951 1952 1953 1954
AFG Afghanistan 20,249 21,352 22,532 23,557 24,555
ALB Albania 8,097 8,986 10,058 11,123 12,246"), header=TRUE)
library(reshape)
x2 <- melt(x, id = c("Code", "Country"), variable_name = "Year")
x2[,"Year"] <- as.numeric(gsub("X", "" , x2[,"Year"]))
由於這個答案被標記為r-faq ,我覺得從基礎 R 分享另一個替代方案會很有用: stack
。
但是請注意,該stack
不適用於factor
s——它僅在is.vector
為TRUE
時才有效,並且從is.vector
的文檔中,我們發現:
如果 x 是指定模式的向量,除了 names 之外沒有其他屬性,
is.vector
返回TRUE
。 否則返回FALSE
。
我正在使用來自@Jaap's answer的樣本數據,其中年份列中的值是factor
s。
這是stack
方法:
cbind(wide[1:2], stack(lapply(wide[-c(1, 2)], as.character)))
## Code Country values ind
## 1 AFG Afghanistan 20,249 1950
## 2 ALB Albania 8,097 1950
## 3 AFG Afghanistan 21,352 1951
## 4 ALB Albania 8,986 1951
## 5 AFG Afghanistan 22,532 1952
## 6 ALB Albania 10,058 1952
## 7 AFG Afghanistan 23,557 1953
## 8 ALB Albania 11,123 1953
## 9 AFG Afghanistan 24,555 1954
## 10 ALB Albania 12,246 1954
這是另一個示例,展示了使用tidyr
gather
的方法。 您可以選擇要gather
的列,方法是單獨刪除它們(就像我在這里所做的那樣),或者明確包含您想要的年份。
請注意,為了處理逗號(如果未設置check.names = FALSE
則添加 X),我還使用dplyr
的 mutate 和parse_number
中的readr
將文本值轉換回數字。 這些都是tidyverse
的一部分,因此可以與library(tidyverse)
一起加載
wide %>%
gather(Year, Value, -Code, -Country) %>%
mutate(Year = parse_number(Year)
, Value = parse_number(Value))
回報:
Code Country Year Value
1 AFG Afghanistan 1950 20249
2 ALB Albania 1950 8097
3 AFG Afghanistan 1951 21352
4 ALB Albania 1951 8986
5 AFG Afghanistan 1952 22532
6 ALB Albania 1952 10058
7 AFG Afghanistan 1953 23557
8 ALB Albania 1953 11123
9 AFG Afghanistan 1954 24555
10 ALB Albania 1954 12246
這是一個sqldf解決方案:
sqldf("Select Code, Country, '1950' As Year, `1950` As Value From wide
Union All
Select Code, Country, '1951' As Year, `1951` As Value From wide
Union All
Select Code, Country, '1952' As Year, `1952` As Value From wide
Union All
Select Code, Country, '1953' As Year, `1953` As Value From wide
Union All
Select Code, Country, '1954' As Year, `1954` As Value From wide;")
要在不輸入所有內容的情況下進行查詢,您可以使用以下命令:
感謝 G. Grothendieck 實施它。
ValCol <- tail(names(wide), -2)
s <- sprintf("Select Code, Country, '%s' As Year, `%s` As Value from wide", ValCol, ValCol)
mquery <- paste(s, collapse = "\n Union All\n")
cat(mquery) #just to show the query
#> Select Code, Country, '1950' As Year, `1950` As Value from wide
#> Union All
#> Select Code, Country, '1951' As Year, `1951` As Value from wide
#> Union All
#> Select Code, Country, '1952' As Year, `1952` As Value from wide
#> Union All
#> Select Code, Country, '1953' As Year, `1953` As Value from wide
#> Union All
#> Select Code, Country, '1954' As Year, `1954` As Value from wide
sqldf(mquery)
#> Code Country Year Value
#> 1 AFG Afghanistan 1950 20,249
#> 2 ALB Albania 1950 8,097
#> 3 AFG Afghanistan 1951 21,352
#> 4 ALB Albania 1951 8,986
#> 5 AFG Afghanistan 1952 22,532
#> 6 ALB Albania 1952 10,058
#> 7 AFG Afghanistan 1953 23,557
#> 8 ALB Albania 1953 11,123
#> 9 AFG Afghanistan 1954 24,555
#> 10 ALB Albania 1954 12,246
不幸的是,我認為PIVOT
和UNPIVOT
不適用於R
SQLite
。 如果您想以更復雜的方式編寫查詢,還可以查看以下帖子:
也可以使用cdata
包,它使用(轉換)控制表的概念:
# data
wide <- read.table(text="Code Country 1950 1951 1952 1953 1954
AFG Afghanistan 20,249 21,352 22,532 23,557 24,555
ALB Albania 8,097 8,986 10,058 11,123 12,246", header=TRUE, check.names=FALSE)
library(cdata)
# build control table
drec <- data.frame(
Year=as.character(1950:1954),
Value=as.character(1950:1954),
stringsAsFactors=FALSE
)
drec <- cdata::rowrecs_to_blocks_spec(drec, recordKeys=c("Code", "Country"))
# apply control table
cdata::layout_by(drec, wide)
我目前正在探索該軟件包並發現它很容易獲得。 它是為更復雜的轉換而設計的,包括反向轉換。 有一個教程可用。
您還可以在R 食譜中看到許多示例
olddata_wide <- read.table(header=TRUE, text='
subject sex control cond1 cond2
1 M 7.9 12.3 10.7
2 F 6.3 10.6 11.1
3 F 9.5 13.1 13.8
4 M 11.5 13.4 12.9
')
# Make sure the subject column is a factor
olddata_wide$subject <- factor(olddata_wide$subject)
olddata_long <- read.table(header=TRUE, text='
subject sex condition measurement
1 M control 7.9
1 M cond1 12.3
1 M cond2 10.7
2 F control 6.3
2 F cond1 10.6
2 F cond2 11.1
3 F control 9.5
3 F cond1 13.1
3 F cond2 13.8
4 M control 11.5
4 M cond1 13.4
4 M cond2 12.9
')
# Make sure the subject column is a factor
olddata_long$subject <- factor(olddata_long$subject)
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