calculate distances between different trips in loops

Question

I have a data frame called p.1 with 16 column of information for each of the points recorded by the Argos system, this are the important columns, date, lon (longitud), lat (latitud), trip is the number of the trip and the distance to the coast.

date               |   lon        |    lat        | trip |  distancetocoast
2014-06-11 19:02:00 -58,3508585     -51,88438373    1       2236,067977
2014-06-12 01:02:00 -58,35589725    -51,88349529    1       1000
2014-06-12 13:02:00 -58,27224941    -51,91903677    1       7211,102551
2014-06-12 19:02:00 -58,27974654    -51,90535003    1       5830,951895
2014-06-13 01:02:00 -58,32331901    -51,89410464    1       3605,551275
2014-06-13 07:02:00 -58,35833139    -51,88809227    1       1414,213562
2014-06-13 13:02:00 -58,35617673    -51,88156281    1       1000
2014-06-13 19:02:00 -58,34055711    -51,89002367    1       2236,067977
2014-06-14 01:02:00 -58,34982536    -51,8715761     2       1000
2014-06-14 13:02:00 -58,3073814     -51,92722937    2       7071,067812
2014-06-14 19:02:00 -58,34581314    -51,86761133    3       1000
2014-06-15 01:02:00 -58,34050624    -51,88382088    3       1414,213562
2014-06-15 13:02:00 -58,2974691     -51,91795326    3       6324,55532
2014-06-15 19:02:00 -58,19881901    -51,95172233    3       13000
2014-06-16 01:02:00 -58,1348416     -51,98673766    3       18788,29423
2014-06-16 07:02:00 -57,99399544    -52,06988191    3       28861,73938
2014-06-16 13:02:00 -58,00469754    -52,02795069    3       26627,05391
2014-06-16 19:02:00 -57,92758675    -52,02184666    3       29000
2014-06-17 01:02:00 -57,91658235    -51,99748699    3       28284,27125
2014-06-17 07:02:00 -57,77015528    -51,99031797    3       30805,8436
2014-06-17 13:02:00 -57,99601712    -51,91519551    3       17804,49381
2014-06-17 19:02:00 -58,06820013    -51,92972737    3       14866,06875
2014-06-18 01:02:00 -58,19845185    -51,89522513    3       7615,773106
2014-06-18 07:02:00 -58,35241361    -51,88015998    3       1000
2014-06-18 13:02:00 -58,35603546    -51,88336878    3       1000
2014-06-18 19:02:00 -58,33350332    -51,87308427    3       1000
2014-06-19 01:02:00 -58,33839581    -51,87846631    3       1414,213562
2014-06-19 07:02:00 -58,42661519    -51,80902388    4       0
2014-06-19 13:02:00 -58,30461883    -51,93745837    4       7810,249676
2014-06-19 19:02:00 -58,18362875    -51,96475914    4       14317,82106

In the beggining I just sumed the distance from coast of every trip with the next function:

tripID<-c(1:length(unique(p.1$trip)))
tripdistance<-aggregate(p.1$dist,by=list(unique.values=p.1$trip),FUN=sum)

This is clearly wrong, how can I calculate the distance traveled in each trip and start from 0 in the next trip?

I´ve been trying with other functions as: spDistsN1 and trackDistance but I can´t make it work.

score 0 · Answer 1 · answered Jun 03 '17 at 16:49

0

Add another column with distance between current row lat long and next row lat long. Put this in a if condition to check the trip number. Finally sum-it up.

First do this in excel and then transfer to your coding language.

answered Jun 03 '17 at 16:49

Sriram Nadiminti

107
2
12

Peter · Answer 2 · 2017-06-04T14:52:54.443

You may find the dplyr to provide the tools you need to do this.

First, if you plan on posting more questions on SO, it will be helpful to use dput to provide copies of your data set in a way that is easy for others to reproduce. For this example, it was easy enough to read your data set in via readr.

library(readr)
library(dplyr)
library(trip)

p.1 <-
  readr::read_delim(delim = "|",
                    locale = locale(decimal_mark = ",", grouping_mark = "."),
                    trim_ws = TRUE,
                    file = 
"date               | lon          | lat          | trip | distancetocoast
2014-06-11 19:02:00 | -58,3508585  | -51,88438373 | 1    | 2236,067977
2014-06-12 01:02:00 | -58,35589725 | -51,88349529 | 1    | 1000
2014-06-12 13:02:00 | -58,27224941 | -51,91903677 | 1    | 7211,102551
2014-06-12 19:02:00 | -58,27974654 | -51,90535003 | 1    | 5830,951895
2014-06-13 01:02:00 | -58,32331901 | -51,89410464 | 1    | 3605,551275
2014-06-13 07:02:00 | -58,35833139 | -51,88809227 | 1    | 1414,213562
2014-06-13 13:02:00 | -58,35617673 | -51,88156281 | 1    | 1000
2014-06-13 19:02:00 | -58,34055711 | -51,89002367 | 1    | 2236,067977
2014-06-14 01:02:00 | -58,34982536 | -51,8715761  | 2    | 1000
2014-06-14 13:02:00 | -58,3073814  | -51,92722937 | 2    | 7071,067812
2014-06-14 19:02:00 | -58,34581314 | -51,86761133 | 3    | 1000
2014-06-15 01:02:00 | -58,34050624 | -51,88382088 | 3    | 1414,213562
2014-06-15 13:02:00 | -58,2974691  | -51,91795326 | 3    | 6324,55532
2014-06-15 19:02:00 | -58,19881901 | -51,95172233 | 3    | 13000
2014-06-16 01:02:00 | -58,1348416  | -51,98673766 | 3    | 18788,29423
2014-06-16 07:02:00 | -57,99399544 | -52,06988191 | 3    | 28861,73938
2014-06-16 13:02:00 | -58,00469754 | -52,02795069 | 3    | 26627,05391
2014-06-16 19:02:00 | -57,92758675 | -52,02184666 | 3    | 29000
2014-06-17 01:02:00 | -57,91658235 | -51,99748699 | 3    | 28284,27125
2014-06-17 07:02:00 | -57,77015528 | -51,99031797 | 3    | 30805,8436
2014-06-17 13:02:00 | -57,99601712 | -51,91519551 | 3    | 17804,49381
2014-06-17 19:02:00 | -58,06820013 | -51,92972737 | 3    | 14866,06875
2014-06-18 01:02:00 | -58,19845185 | -51,89522513 | 3    | 7615,773106
2014-06-18 07:02:00 | -58,35241361 | -51,88015998 | 3    | 1000
2014-06-18 13:02:00 | -58,35603546 | -51,88336878 | 3    | 1000
2014-06-18 19:02:00 | -58,33350332 | -51,87308427 | 3    | 1000
2014-06-19 01:02:00 | -58,33839581 | -51,87846631 | 3    | 1414,213562
2014-06-19 07:02:00 | -58,42661519 | -51,80902388 | 4    | 0
2014-06-19 13:02:00 | -58,30461883 | -51,93745837 | 4    | 7810,249676
2014-06-19 19:02:00 | -58,18362875 | -51,96475914 | 4    | 14317,82106")

dplyr provides a group_by function that will do exactly what it sounds like, create groups defined by the values within one or more columns. The mutate command creates columns of a data set.

A vector distance_traveled has been created to start at 0 for the first entry of each trip and then use spDist for the distance traveled.

The cumulative distance and the total distance for each trip is also reported. If you are unfamiliar with the %>% operator, read help("%>%", package = "magrittr").

p.1 %>%
group_by(trip) %>%
arrange(date) %>% 
mutate(distance_traveled       = c(0, spDists(x = cbind(lon, lat), longlat = TRUE, segments = TRUE)),
       cumlative_trip_distance = cumsum(distance_traveled),
       total_trip_distance     = sum(distance_traveled)) %>%
ungroup() %>%
print.data.frame

p.1
#                   date       lon       lat trip distancetocoast distance_traveled cumlative_trip_distance total_trip_distance
# 1  2014-06-11 19:02:00 -58.35086 -51.88438    1        2236.068         0.0000000               0.0000000           16.875313
# 2  2014-06-12 01:02:00 -58.35590 -51.88350    1        1000.000         0.3607526               0.3607526           16.875313
# 3  2014-06-12 13:02:00 -58.27225 -51.91904    1        7211.103         6.9846980               7.3454506           16.875313
# 4  2014-06-12 19:02:00 -58.27975 -51.90535    1        5830.952         1.6078870               8.9533376           16.875313
# 5  2014-06-13 01:02:00 -58.32332 -51.89410    1        3605.551         3.2496828              12.2030205           16.875313
# 6  2014-06-13 07:02:00 -58.35833 -51.88809    1        1414.214         2.5015138              14.7045343           16.875313
# 7  2014-06-13 13:02:00 -58.35618 -51.88156    1        1000.000         0.7415006              15.4460348           16.875313
# 8  2014-06-13 19:02:00 -58.34056 -51.89002    1        2236.068         1.4292778              16.8753126           16.875313
# 9  2014-06-14 01:02:00 -58.34983 -51.87158    2        1000.000         0.0000000               0.0000000            6.846905
# 10 2014-06-14 13:02:00 -58.30738 -51.92723    2        7071.068         6.8469049               6.8469049            6.846905
# 11 2014-06-14 19:02:00 -58.34581 -51.86761    3        1000.000         0.0000000               0.0000000          103.020176
# 12 2014-06-15 01:02:00 -58.34051 -51.88382    3        1414.214         1.8402281               1.8402281          103.020176
# 13 2014-06-15 13:02:00 -58.29747 -51.91795    3        6324.555         4.8164202               6.6566483          103.020176
# 14 2014-06-15 19:02:00 -58.19882 -51.95172    3       13000.000         7.7558235              14.4124718          103.020176
# 15 2014-06-16 01:02:00 -58.13484 -51.98674    3       18788.294         5.8746536              20.2871254          103.020176
# 16 2014-06-16 07:02:00 -57.99400 -52.06988    3       28861.739        13.3804466              33.6675720          103.020176
# 17 2014-06-16 13:02:00 -58.00470 -52.02795    3       26627.054         4.7230685              38.3906405          103.020176
# 18 2014-06-16 19:02:00 -57.92759 -52.02185    3       29000.000         5.3362603              43.7269008          103.020176
# 19 2014-06-17 01:02:00 -57.91658 -51.99749    3       28284.271         2.8138115              46.5407123          103.020176
# 20 2014-06-17 07:02:00 -57.77016 -51.99032    3       30805.844        10.0892378              56.6299501          103.020176
# 21 2014-06-17 13:02:00 -57.99602 -51.91520    3       17804.494        17.6348017              74.2647518          103.020176
# 22 2014-06-17 19:02:00 -58.06820 -51.92973    3       14866.069         5.2225348              79.4872866          103.020176
# 23 2014-06-18 01:02:00 -58.19845 -51.89523    3        7615.773         9.7503783              89.2376649          103.020176
# 24 2014-06-18 07:02:00 -58.35241 -51.88016    3        1000.000        10.7319100              99.9695749          103.020176
# 25 2014-06-18 13:02:00 -58.35604 -51.88337    3        1000.000         0.4355097             100.4050846          103.020176
# 26 2014-06-18 19:02:00 -58.33350 -51.87308    3        1000.000         1.9279737             102.3330583          103.020176
# 27 2014-06-19 01:02:00 -58.33840 -51.87847    3        1414.214         0.6871181             103.0201763          103.020176
# 28 2014-06-19 07:02:00 -58.42662 -51.80902    4           0.000         0.0000000               0.0000000           25.433053
# 29 2014-06-19 13:02:00 -58.30462 -51.93746    4        7810.250        16.5773793              16.5773793           25.433053
# 30 2014-06-19 19:02:00 -58.18363 -51.96476    4       14317.821         8.8556734              25.4330527           25.433053

The total distance travelled by the penguin in each trip is not the sum of the distance from the coast, is the distance from the first point to the next+the distance with the next and successively. What I did to calculate the distance between the different points was: `distancebetwenpoints=spDists(locs1_utm, longlat=FALSE)` `p.1$dist=distancebetwenpoints ` `locs1_utm$dist=distancebetwenpoints` locs1_utm is the same dataframe p.1 but in format SpatialPointsDataFrame. This give the distance between points, I need to reset that for each new trip and sum the distances in each trip. — Daniel González, Jun 04 '17 at 09:44
The problem i have as well is that with `distancebetwenpoints=spDists(locs1_utm, longlat=FALSE)` is that it doesn´t restart the distance everytime that starts a new trip — Daniel González, Jun 04 '17 at 09:53
For some reason I am trying to run the code but R send the next message: _Error: incompatible size (2), expecting 1 (the group size) or 1_ I am not sure where can be the problem — Daniel González, Jun 04 '17 at 15:43
Hi @Peter, I can´t make work the code, every time i try to work it out I have the same message as above, _Error: incompatible size (2), expecting 1 (the group size) or 1_ you know where it could be the problem? using the spDist it creates a matrix but I can´t run statistics with the matrix properly — Daniel González, Jun 06 '17 at 15:26
I have use this: `distancebetwenpoints=spDists(locs1_utm, longlat=TRUE) #this creates a matrix I want a frame p.1$dist=as.factordistancebetwenpoints And using this simple loop I have the cumulative distance `for( i in 1:nrow(p.1)){ if (p.1$todelete[i]==1){ p.1$dist[i]<-0 }else{p.1$dist[i]<-sum(p.1$dist[i]+p.1$dist[i-1])} } ` This give me a huge matrix, within the dataframe, but p.1$dist have the values that i need in one column but I can´t work out with that matrix to analyse it. — Daniel González, Jun 06 '17 at 15:51
I uploaded s shorter version of the excel [https://drive.google.com/open?id=0B7mltLkUahnQM0RjMDRkOGFDc0U] @Peter — Daniel González, Jun 07 '17 at 11:10

calculate distances between different trips in loops

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