[英]R - Calculate the differences in the column values between rows/ observations (all combinations)
[英]How to calculate all possible combinations of distance and time differences between grouped rows of data in R?
我的數據包括美元鈔票的距離和時間。 我的數據如下:
bid ts latitude longitude
1 123 0 38.40513 41.83777
2 123 23 38.41180 41.68493
3 123 45 42.20771 43.36318
4 123 50 40.22803 43.00208
5 456 0 39.12882 42.73877
6 456 12 38.46078 42.79847
7 456 27 40.53698 42.57617
8 456 19 39.04038 42.17070
9 234 0 39.18274 41.17445
10 234 8 39.58652 43.61317
11 234 15 41.32383 41.49377
12 234 23 40.26008 42.01927
bid = bill id
ts =當t = 0時從原始數據點計算的時間戳(天)
緯度和經度=位置
此數據顯示了美國各地的賬單ID的變動情況。
我想計算每個類似行組4的所有可能組合之間的平方距離和時間差。例如,對於出價組123我想計算距離和時間之間的差異:第1行和第2行,第1行和第3行,第1行和第4行,第2行和第3行,第2行和第4行,第3行和第4行。
這將為我提供這一組投標之間所有可能的計算組合。
我能夠在連續行之間使用dplyr執行此操作,如下所示:
detach("package:plyr", unload=TRUE)
library(magrittr)
library(dplyr)
library(geosphere)
deltadata <- group_by(df, bid) %>%
mutate(
dsq = (c(NA,distHaversine(cbind(longitude[-n()], latitude[-n()]),
cbind(longitude[ -1], latitude[ -1]))))^2,
dt = c(NA, diff(ts))
)%>%
ungroup() %>%
filter( ! is.na(dsq) )
deltadata
# A tibble: 21 x 6
bid ts latitude longitude dsq dt
<dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 123 23 38.41180 41.68493 178299634 23
2 123 45 42.20771 43.36318 198827672092 22
3 123 50 40.22803 43.00208 49480260636 5
4 456 12 38.46078 42.79847 5557152213 12
5 456 27 40.53698 42.57617 53781504422 15
6 456 19 39.04038 42.17070 28958550947 -8
7 234 8 39.58652 43.61317 46044153364 8
8 234 15 41.32383 41.49377 69621429008 7
9 234 23 40.26008 42.01927 15983792199 8
10 345 5 40.25700 41.69525 26203255328 5
# ... with 11 more rows
問題:這只計算連續行之間的平方距離和時間,即:第1行和第2行,第2行和第3行,第3行和第4行
有沒有一種實用的方法可以為每組中所有可能的行組合執行此操作?
我希望我的輸出對每個出價有6個計算,如下所示:
# A tibble: 21 x 6
bid ts latitude longitude dsq dt
<dbl> <dbl> <dbl> <dbl> <dbl> <dbl>
1 123 23 38.41180 41.68493 178299634 23 (for rows 1 and 2)
2 123 45 42.20771 43.36318 198827672092 22 (for rows 1 and 3)
3 123 50 40.22803 43.00208 49480260636 5 (for rows 1 and 4)
4 123 12 38.46078 42.79847 5557152213 12 (for rows 2 and 3)
5 123 27 40.53698 42.57617 53781504422 15 (for rows 2 and 4)
6 123 19 39.04038 42.17070 28958550947 -8 (for rows 2 and 5)
我是R的新手,所以任何建議都表示贊賞!
你可以像這樣使用inner_join
:
library(dplyr)
library(geosphere)
df <- read.table(text = ' bid ts latitude longitude
1 123 0 38.40513 41.83777
2 123 23 38.41180 41.68493
3 123 45 42.20771 43.36318
4 123 50 40.22803 43.00208
5 456 0 39.12882 42.73877
6 456 12 38.46078 42.79847
7 456 27 40.53698 42.57617
8 456 19 39.04038 42.17070
9 234 0 39.18274 41.17445
10 234 8 39.58652 43.61317
11 234 15 41.32383 41.49377
12 234 23 40.26008 42.01927')
df %>%
inner_join(df, by = c("bid" = "bid")) %>%
mutate(
dsq = distHaversine(cbind(longitude.x, latitude.x),
cbind(longitude.y, latitude.y))^2,
dt = ts.x -ts.y
) %>%
filter(dt > 0)
#> bid ts.x latitude.x longitude.x ts.y latitude.y longitude.y dsq dt
#> 1 123 23 38.41180 41.68493 0 38.40513 41.83777 178300279 23
#> 2 123 45 42.20771 43.36318 0 38.40513 41.83777 195932999496 45
#> 3 123 45 42.20771 43.36318 23 38.41180 41.68493 198827439286 22
#> 4 123 50 40.22803 43.00208 0 38.40513 41.83777 51230447939 50
#> 5 123 50 40.22803 43.00208 23 38.41180 41.68493 53740739037 27
#> 6 123 50 40.22803 43.00208 45 42.20771 43.36318 49479978030 5
#> 7 456 12 38.46078 42.79847 0 39.12882 42.73877 5557111219 12
#> 8 456 27 40.53698 42.57617 0 39.12882 42.73877 24765506646 27
#> 9 456 27 40.53698 42.57617 12 38.46078 42.79847 53781664569 15
#> 10 456 27 40.53698 42.57617 19 39.04038 42.17070 28958542352 8
#> 11 456 19 39.04038 42.17070 0 39.12882 42.73877 2506329323 19
#> 12 456 19 39.04038 42.17070 12 38.46078 42.79847 7133122323 7
#> 13 234 8 39.58652 43.61317 0 39.18274 41.17445 46043956815 8
#> 14 234 15 41.32383 41.49377 0 39.18274 41.17445 57544071797 15
#> 15 234 15 41.32383 41.49377 8 39.58652 43.61317 69621225065 7
#> 16 234 23 40.26008 42.01927 0 39.18274 41.17445 19614888600 23
#> 17 234 23 40.26008 42.01927 8 39.58652 43.61317 24136886438 15
#> 18 234 23 40.26008 42.01927 15 41.32383 41.49377 15983645507 8
既然你也使用了data.table標簽,這里有一個使用該軟件包的解決方案:
library(data.table)
library(geosphere)
df <- read.table(text = ' bid ts latitude longitude
1 123 0 38.40513 41.83777
2 123 23 38.41180 41.68493
3 123 45 42.20771 43.36318
4 123 50 40.22803 43.00208
5 456 0 39.12882 42.73877
6 456 12 38.46078 42.79847
7 456 27 40.53698 42.57617
8 456 19 39.04038 42.17070
9 234 0 39.18274 41.17445
10 234 8 39.58652 43.61317
11 234 15 41.32383 41.49377
12 234 23 40.26008 42.01927')
dt <- data.table(df, key = 'bid')
dt <- dt[dt, allow.cartesian = TRUE][ts < i.ts]
dt[, dt := i.ts - ts][, dsq := distHaversine(cbind(longitude, latitude),
cbind(i.longitude, i.latitude))^2]
dt
#> bid ts latitude longitude i.ts i.latitude i.longitude dt dsq
#> 1: 123 0 38.40513 41.83777 23 38.41180 41.68493 23 178300279
#> 2: 123 0 38.40513 41.83777 45 42.20771 43.36318 45 195932999496
#> 3: 123 23 38.41180 41.68493 45 42.20771 43.36318 22 198827439286
#> 4: 123 0 38.40513 41.83777 50 40.22803 43.00208 50 51230447939
#> 5: 123 23 38.41180 41.68493 50 40.22803 43.00208 27 53740739037
#> 6: 123 45 42.20771 43.36318 50 40.22803 43.00208 5 49479978030
#> 7: 234 0 39.18274 41.17445 8 39.58652 43.61317 8 46043956815
#> 8: 234 0 39.18274 41.17445 15 41.32383 41.49377 15 57544071797
#> 9: 234 8 39.58652 43.61317 15 41.32383 41.49377 7 69621225065
#> 10: 234 0 39.18274 41.17445 23 40.26008 42.01927 23 19614888600
#> 11: 234 8 39.58652 43.61317 23 40.26008 42.01927 15 24136886438
#> 12: 234 15 41.32383 41.49377 23 40.26008 42.01927 8 15983645507
#> 13: 456 0 39.12882 42.73877 12 38.46078 42.79847 12 5557111219
#> 14: 456 0 39.12882 42.73877 27 40.53698 42.57617 27 24765506646
#> 15: 456 12 38.46078 42.79847 27 40.53698 42.57617 15 53781664569
#> 16: 456 19 39.04038 42.17070 27 40.53698 42.57617 8 28958542352
#> 17: 456 0 39.12882 42.73877 19 39.04038 42.17070 19 2506329323
#> 18: 456 12 38.46078 42.79847 19 39.04038 42.17070 7 7133122323
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