Question

我正在使用两个不同的数据框。在第一个中，我有地理质心（点）的行名称，以及另一组点（火车站）的列名ID。这些值表示质心与列中的点之间的距离（由gDistance提取）。

从这一点开始，我提取了3个最接近的点，在翻译的数据框中保存了它们的ID。这个现在作为行的质心ID和列只是等级（1,2,3）。

distances<- ft2miles(gDistance(stations, centroids, byid=TRUE))
access <-as.data.frame(apply(distances, 1, function(X) colnames(distances)[order(X)][1:3]))
access <- as.data.frame(t(access))

我想在最后一个数据框中添加三个具有相应距离值的列。更清楚的是，距离数据框的列对应于第二列的值，行名称是相同的。

第一个数据框：

                   158      1559     1560     1561      1722      1723     1732 166516384      1546
croatia_new.0 1.108585 0.9434521 1.138825 1.621118 0.8457217 0.5554764 2.441048 0.7393321 6.4736425
croatia_new.1 1.107511 1.3381573 2.970976 3.448820 2.6142722 2.0175318 4.167618 1.6984548 6.2927608

第二个数据框：

              closest1  closest2  closest3
croatia_new.0     1723 166516384      1722
croatia_new.1      158      1559 166516384
croatia_new.2     1546      1584  98349400
croatia_new.3     1584      1546  98349400

该数据帧的值基本上是第一个数据帧的列名，表示最近站的ID。我想要获得什么：

                closest1    closest2    closest3
croatia_new.0   0.5554764   0.739332    0.8457217
croatia_new.1   1.107511    1.3381573   1.6984548
croatia_new.N    ....         ....       ....

使用DigEmAll解决方案获得的输出

                     V1       V2       V3
croatia_new.0  3.719507       NA       NA
croatia_new.1 63.082290 38.04791 36.94511
croatia_new.2        NA       NA 49.05636
croatia_new.3 55.836949 47.74032 46.48799
croatia_new.4 41.484741       NA 52.85941
croatia_new.5 60.654512       NA 49.68948

控制台输出：

> dput(distances["croatia_new.2",1:10])
structure(list(`158` = 6.99119002871976, `1559` = 7.05366412218041, 
`1560` = 6.99165462844427, `1561` = 6.79651949339854, `1722` =     7.09268368796389, 
    `1723` = 7.06552376797322, `1732` = 6.3722523477356, `166516384` = 7.1279696005328, 
`1546` = 0.153911810037831, `1584` = 0.433837036180278), .Names = c("158", 
"1559", "1560", "1561", "1722", "1723", "1732", "166516384", 
"1546", "1584"), row.names = "croatia_new.2", class = "data.frame")

dput(access["croatia_new.2",])

structure(list(V1 = structure(149L, .Names = "croatia_new.2", .Label =    c("1001", 
"1002", "1003", "1004", "1005", "1006", "1007", "1018", "1019", 
"1020", "1023", "1024", "1025", "1031", "1038", "1042", "1045", 
"1046", "1047", "1048", "1050", "1052", "1053", "1054", "1056", 
"105753770", "1059", "1061", "1062", "1064", "1065", "1067", 
"1073", "1076", "1085", "1088", "109", "1091", "1093", "1095", ....."999"),   class = "factor")), .Names = c("V1", 
"V2", "V3"), row.names = "croatia_new.2", class = "data.frame")

Answer 1

我想你想要这样的东西：

### RECREATE THE INPUT EXAMPLE
DF1 <- read.csv(text=
"158,1559,1560,1561,1722,1723,1732,166516384,1546
croatia_new.0,1.108585,0.9434521,1.138825,1.621118,0.8457217,0.5554764,2.441048,0.7393321,6.4736425
croatia_new.1,1.107511,1.3381573,2.970976,3.448820,2.6142722,2.0175318,4.167618,1.6984548,6.2927608",
check.names=FALSE
)
DF2 <- read.csv(text=
"closest1,closest2,closest3
croatia_new.0,1723,166516384,1722
croatia_new.1,158,1559,166516384
croatia_new.2,1546,1584,98349400
croatia_new.3,1584,1546,98349400"
)

# ensure the values of DF2 are characters
DF2[] <- lapply(DF2, as.character)

# for each row of DF2, we search for the value in DF1 using the
# row names and column names stored in DF2.
# The result will be a list of numeric values.
rows2bind <- 
  lapply(1:nrow(DF2),function(i){
    station <- row.names(DF2)[i]
    row <- DF2[i,]
    row2 <- as.numeric(DF1[match(station,row.names(DF1)),])
    return(row2[match(as.character(row),colnames(DF1))])
  })

# we turn the list into a data.frame similar to DF2 but with values instead of column names
DF2ext <- setNames(as.data.frame(do.call(rbind,rows2bind),row.names=row.names(DF2)), colnames(DF2))

# we add the closest columns to DF1 
DF1ext <- cbind(DF1,DF2ext[match(row.names(DF1),row.names(DF2ext)),])

结果：
（注意DF1ext的最后3列）

> DF2ext
               closest1  closest2  closest3
croatia_new.0 0.5554764 0.7393321 0.8457217
croatia_new.1 1.1075110 1.3381573 1.6984548
croatia_new.2        NA        NA        NA
croatia_new.3        NA        NA        NA

> DF1ext
                   158      1559     1560     1561      1722      1723     1732 166516384     1546  closest1  closest2  closest3
croatia_new.0 1.108585 0.9434521 1.138825 1.621118 0.8457217 0.5554764 2.441048 0.7393321 6.473643 0.5554764 0.7393321 0.8457217
croatia_new.1 1.107511 1.3381573 2.970976 3.448820 2.6142722 2.0175318 4.167618 1.6984548 6.292761 1.1075110 1.3381573 1.6984548

根据从第二个数据帧派生的列名，从数据框中提取值

1 个答案: