Question

我使用代码

形成了以下数据

test <- data.frame(dis = c(10,20,30,40),dur=c(30,40,60,90),method=c("car","car","Bicycle","Bicycle"),to_lon=c(-1.980,-1.5678,-1.324,-1.456),to_lat=c(55.3009,55.3416,55.1123,55.2234),from_lon=c(-1.4565,-1.3424,-1.4566,-1.1111),from_lat=c(76.8888,65.8999,76.9088,25.3344))

 dis dur  method  to_lon  to_lat from_lon from_lat
1  10  30     car -1.9800 55.3009  -1.4565  76.8888
2  20  40     car -1.5678 55.3416  -1.3424  65.8999
3  30  60 Bicycle -1.3240 55.1123  -1.4566  76.9088
4  40  90 Bicycle -1.4560 55.2234  -1.1111  25.3344

我想转换这个数据框，使得它有一行to_lat和to_lon，而在下一行它有from_lat和from_lon。其余细节不需要更改，可以复制。期望的结果应如下所示

    dis dur method  longitude   latitude
from    10  30  car -1.98   55.3009
to  10  30  car -1.4565 76.8888
from    20  40  car -1.5678 55.3416
to  20  40  car -1.3424 65.8999
from    30  60  Bicycle -1.324  55.1123
to  30  60  Bicycle -1.4566 76.9088
from    40  90  Bicycle -1.456  55.2234
to  40  90  Bicycle -1.1111 25.3344

非常感谢任何帮助。

感谢。

Answer 1

我们可以使用melt中的data.table，measure可以使用多个library(data.table) dM <- melt(setDT(test), measure=patterns('lon', 'lat'), value.name=c('longitude', 'latitude')) #change the 'variable' column from numeric index to 'from/to' dM[, variable:= c('from', 'to')[variable]] #create a sequence column grouped by 'variable' dM[,i1:= 1:.N ,variable] #order based on the 'i1' res <- dM[order(i1)][,i1:=NULL] res # dis dur method variable longitude latitude #1: 10 30 car from -1.9800 55.3009 #2: 10 30 car to -1.4565 76.8888 #3: 20 40 car from -1.5678 55.3416 #4: 20 40 car to -1.3424 65.8999 #5: 30 60 Bicycle from -1.3240 55.1123 #6: 30 60 Bicycle to -1.4566 76.9088 #7: 40 90 Bicycle from -1.4560 55.2234 #8: 40 90 Bicycle to -1.1111 25.3344列。

df['Merged'] = df['Questions'] + df['Answers']

Answer 2

这可能不是最优雅的解决方案，但它应该有效并且希望可以理解：

我们将数据分成两个数据帧：一个带有'from'经度和纬度数据（称之为testF），另一个带有'to'数据（称之为测试）。然后我们使用rbind将'testF'行插入'test'中的适当位置。

test <- data.frame(dis = c(10,20,30,40),dur=c(30,40,60,90),method=c("car","car","Bicycle","Bicycle"),to_lon=c(-1.980,-1.5678,-1.324,-1.456),to_lat=c(55.3009,55.3416,55.1123,55.2234),from_lon=c(-1.4565,-1.3424,-1.4566,-1.1111),from_lat=c(76.8888,65.8999,76.9088,25.3344))

testF <- test[,c(1:3,6,7)]
names(testF)[4:5] <- c("lonitude", "latitude")
test <- test[,1:5]
names(test)[4:5] <- c("lonitude", "latitude")

for(i in dim(test)[1]:1) {
  test <- rbind(test[1:i,], testF[i,], test[-(1:i),])
}

Answer 3

以下是使用包tidyr（一种流行的数据保持包）的替代方法，它避免了for循环。

library(tidyr)

test <- data.frame(dis = c(10,20,30,40),dur=c(30,40,60,90),method=c("car","car","Bicycle","Bicycle"),to_lon=c(-1.980,-1.5678,-1.324,-1.456),to_lat=c(55.3009,55.3416,55.1123,55.2234),from_lon=c(-1.4565,-1.3424,-1.4566,-1.1111),from_lat=c(76.8888,65.8999,76.9088,25.3344))
test$id <- 1:dim(test)[1]

# gather latitude columns
d1 <- gather(data = test, 
             key = direction, 
             value = latitude, 
             to_lat, from_lat)

# gather longitude columns
d2 <- gather(data = test, 
             key = direction, 
             value = longitude, 
             to_lon, from_lon)

d3 <- cbind(d1[,c("direction","dis","dur","method","latitude")],d2[,c("longitude","id"),drop=FALSE])

# Create names
dir <- unlist(strsplit(d3$direction,"_"))
dir <- dir[seq(from = 1, to = length(dir), by = 2)]

# Factor and sort
d3$direction <- factor(dir)
d3[order(d3$id),]

将列转换为r

3 个答案: