Convert data frame to spatial lines data frame in R with x,y x,y coordintates

问题

I have a data frame in R, one of the columns contains the coordinates for points along a line in the form:

x,y x,y x,y x,y

So the whole data frame looks like

id dist speed coord
1  45   6     1.294832,54.610240 -1.294883,54.610080 -1.294262,54.6482757
2  23   34    2.788732,34.787940 6.294883,24.567080 -5.564262,-45.7676757

I would like to convert this to a spatial lines data frame, and I assume that the fist step would be to separate the coordinates into two columns in the from:

x, x, x, x
y, y, y, y

But I am unsure how to proceed.

EDIT

For those requesting it a dput of the actual file

> finalsub <- final[final$rid <3,]
> dput(finalsub)
structure(list(rid = c(1, 2), start_id = c(1L, 1L), start_code = c("E02002536", 
"E02002536"), end_id = c(106L, 106L), end_code = c("E02006909", 
"E02006909"), strategy = c("fastest", "quietest"), distance = c(12655L, 
12909L), time_seconds = c(2921L, 3422L), calories = c(211L, 201L
), document.id = c(1L, 1L), array.index = 1:2, start = c("Geranium Close", 
"Geranium Close"), finish = c("Hylton Road", "Hylton Road"), 
    startBearing = c(0, 0), startSpeed = c(0, 0), start_longitude = c(-1.294832, 
    -1.294832), start_latitude = c(54.610241, 54.610241), finish_longitude = c(-1.249478, 
    -1.249478), finish_latitude = c(54.680691, 54.680691), crow_fly_distance = c(8362, 
    8362), event = c("depart", "depart"), whence = c(1473171787, 
    1473171787), speed = c(20, 20), itinerary = c(419956, 419957
    ), clientRouteId = c(0, 0), plan = c("fastest", "quietest"
    ), note = c("", ""), length = c(12655, 12909), time = c(2921, 
    3422), busynance = c(42172, 17242), quietness = c(30, 75), 
    signalledJunctions = c(3, 4), signalledCrossings = c(2, 0
    ), west = c(-1.300074, -1.294883), south = c(54.610006, 54.609851
    ), east = c(-1.232447, -1.232447), north = c(54.683814, 54.683814
    ), name = c("Geranium Close to Hylton Road", "Geranium Close to Hylton Road"
    ), walk = c(0, 0), leaving = c("2016-09-06 15:23:07", "2016-09-06 15:23:07"
    ), arriving = c("2016-09-06 16:11:48", "2016-09-06 16:20:09"
    ), coordinates = c("-1.294832,54.610240 -1.294883,54.610080 -1.294262,54.610016 -1.294141,54.610006 -1.293710,54.610038 -1.293726,54.610142 -1.293742,54.610247 -1.293510,54.610262 -1.293368,54.610258 -1.292816,54.610195 -1.292489,54.610152 -1.292298,54.610667 -1.292205,54.610951 -1.292182,54.611063 -1.292183,54.611153 -1.292239,54.611341 -1.292305,54.611447 -1.292375,54.611534 -1.292494,54.611639 -1.292739,54.611830 -1.292909,54.611980 -1.293010,54.612107 -1.293111,54.612262 -1.293192,54.612423 -1.293235,54.612546 -1.293267,54.612684 -1.293279,54.612818 -1.293510,54.612813 -1.293732,54.612790 -1.294324,54.612691 -1.295086,54.612568 -1.295313,54.612539 -1.295379,54.612543 -1.295889,54.612645 -1.295945,54.612648 -1.296006,54.612642 -1.297154,54.612414 -1.297502,54.612895 -1.297733,54.612847 -1.297990,54.612796 -1.298292,54.612747 -1.298515,54.612727 -1.299088,54.612681 -1.299564,54.612669 -1.299798,54.612663 -1.300006,54.612660 -1.300057,54.612809 -1.300056,54.613335 -1.300071,54.613693 -1.300074,54.614044 -1.300042,54.614482 -1.300015,54.614786 -1.299947,54.615220 -1.299907,54.615394 -1.299854,54.615644 -1.299730,54.616048 -1.299495,54.616700 -1.299196,54.617347 -1.298236,54.619313 -1.298010,54.619762 -1.297703,54.620418 -1.297520,54.620831 -1.297169,54.621690 -1.297061,54.621981 -1.296416,54.623873 -1.296310,54.624308 -1.296225,54.624888 -1.296215,54.625286 -1.296220,54.625546 -1.296241,54.625803 -1.296268,54.625913 -1.296323,54.626011 -1.296397,54.626096 -1.296540,54.626190 -1.296719,54.626323 -1.296893,54.626433 -1.297042,54.626589 -1.297111,54.626710 -1.297122,54.626825 -1.297110,54.626948 -1.297058,54.627052 -1.296961,54.627172 -1.296861,54.627258 -1.296760,54.627325 -1.296603,54.627397 -1.296491,54.627438 -1.296338,54.627472 -1.296154,54.627496 -1.295966,54.627513 -1.295746,54.627526 -1.295618,54.627522 -1.295421,54.627510 -1.295197,54.627466 -1.295102,54.627436 -1.294832,54.627376 -1.294665,54.627355 -1.294502,54.627350 -1.294331,54.627366 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    "-1.294832,54.610240 -1.294883,54.610080 -1.294262,54.610016 -1.294141,54.610006 -1.293710,54.610038 -1.293726,54.610142 -1.293742,54.610247 -1.293510,54.610262 -1.293368,54.610258 -1.292816,54.610195 -1.292489,54.610152 -1.292298,54.610667 -1.292167,54.610651 -1.291371,54.610562 -1.291240,54.610556 -1.291107,54.610564 -1.290983,54.610581 -1.290467,54.610665 -1.290253,54.610690 -1.290017,54.610689 -1.289770,54.610665 -1.289500,54.610620 -1.289281,54.610570 -1.289124,54.610514 -1.288957,54.610440 -1.288611,54.610277 -1.288420,54.610222 -1.287445,54.610110 -1.287259,54.610664 -1.286758,54.610611 -1.285446,54.610462 -1.285308,54.610459 -1.283356,54.610475 -1.283159,54.610475 -1.283156,54.610324 -1.283153,54.610119 -1.282818,54.610118 -1.282560,54.610114 -1.282110,54.610131 -1.281962,54.610153 -1.281788,54.610200 -1.281639,54.610257 -1.281298,54.609964 -1.281196,54.609851 -1.280586,54.610008 -1.280272,54.610054 -1.279816,54.610091 -1.279480,54.610104 -1.279112,54.610121 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-1.234300,54.681119 -1.234362,54.681549 -1.234427,54.681771 -1.234560,54.682172 -1.234782,54.682824 -1.236530,54.682837 -1.236725,54.682829 -1.237133,54.682813 -1.238813,54.683143 -1.241021,54.683814 -1.241819,54.683771 -1.242854,54.683717 -1.242946,54.683718 -1.243082,54.683716 -1.244694,54.683772 -1.244658,54.683077 -1.245038,54.682805 -1.245047,54.681990 -1.245011,54.681238 -1.245220,54.680975 -1.247056,54.680601 -1.248019,54.680404 -1.249478,54.680691"
    ), grammesCO2saved = c(2359, 2406), calories = c(211, 201
    ), type = c("route", "route")), .Names = c("rid", "start_id", 
"start_code", "end_id", "end_code", "strategy", "distance", "time_seconds", 
"calories", "document.id", "array.index", "start", "finish", 
"startBearing", "startSpeed", "start_longitude", "start_latitude", 
"finish_longitude", "finish_latitude", "crow_fly_distance", "event", 
"whence", "speed", "itinerary", "clientRouteId", "plan", "note", 
"length", "time", "busynance", "quietness", "signalledJunctions", 
"signalledCrossings", "west", "south", "east", "north", "name", 
"walk", "leaving", "arriving", "coordinates", "grammesCO2saved", 
"calories", "type"), row.names = 1:2, class = "data.frame")
>

回答1:

I believe what you want to end up with is a column in your data frame that for each row is a list (or data frame) with x.coord and y.coord columns. To achieve that, we can use unnest and nest from tidyr with dplyr:

library(dplyr)
library(tidyr)
result <- finalsub %>% mutate(coordinates = strsplit(coordinates,split=" ",fixed=TRUE)) %>%
                       unnest(coordinates) %>%
                       mutate(coordinates = strsplit(coordinates,split=",",fixed=TRUE),
                              x.coord = as.numeric(unlist(coordinates)[c(TRUE,FALSE)]),
                              y.coord = as.numeric(unlist(coordinates)[c(FALSE,TRUE)])) %>%
                       select(-coordinates) %>%
                       nest(x.coord,y.coord,.key=coordinates)

Notes:

The first mutate splits the character vector in your coordinates column by " " to separate each coordinate x,y resulting in a list of these.
unnest separates this list into rows.
In the second mutate, we first split each coordinate x,y, this time by "," to separate each coordinate into x and y. Then we create separate x.coord and y.coord columns to hold these. Note the conversion to numeric here.
Finally, we use nest to collect the x.coord and y.coord columns as a list under the column named coordinates. Note that we first have to remove the original coordinates column.

The result using your dput data, printing only the coordinates column:

print(result$coordinates)
##[[1]]
### A tibble: 284 x 2
##     x.coord  y.coord
##       <dbl>    <dbl>
##1  -1.294832 54.61024
##2  -1.294883 54.61008
##3  -1.294262 54.61002
##4  -1.294141 54.61001
##5  -1.293710 54.61004
##6  -1.293726 54.61014
##7  -1.293742 54.61025
##8  -1.293510 54.61026
##9  -1.293368 54.61026
##10 -1.292816 54.61019
### ... with 274 more rows
##
##[[2]]
### A tibble: 322 x 2
##     x.coord  y.coord
##       <dbl>    <dbl>
##1  -1.294832 54.61024
##2  -1.294883 54.61008
##3  -1.294262 54.61002
##4  -1.294141 54.61001
##5  -1.293710 54.61004
##6  -1.293726 54.61014
##7  -1.293742 54.61025
##8  -1.293510 54.61026
##9  -1.293368 54.61026
##10 -1.292816 54.61019
### ... with 312 more rows

回答2:

df1 <- data.frame(id= c(1,2), dist =c(45,23), speed = c(6,24) ,do.call(rbind,strsplit(df$cord,split = " ")))

library(reshape2)

df1 <- melt(df1,id=c("id","dist","speed"))

df2<- data.frame(do.call(rbind,strsplit(df1$value, split=",")))
df1$value <- NULL
df1 <- cbind(df1,df2)
names(df1)[5:6] <- c("x","y")
id dist speed variable         x           y
1  1   45     6       X1  1.294832   54.610240
2  2   23    24       X1  2.788732   34.787940
3  1   45     6       X2 -1.294883   54.610080
4  2   23    24       X2  6.294883   24.567080
5  1   45     6       X3 -1.294262  54.6482757
6  2   23    24       X3 -5.564262 -45.7676757

来源：https://stackoverflow.com/questions/39746698/convert-data-frame-to-spatial-lines-data-frame-in-r-with-x-y-x-y-coordintates

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dataframe

spatial-data-frame