CODEWITHSUNDEEP
rjsonstringlistdata.table
jaksco
jaksco

Reputation: 531

Create new column based on strings within the same list in R

Sorry if this is a duplicate. I couldn't find something similar after thirty minutes of searching though.

dput: 

structure(list(i = c("KHH", "KHH", "KHH", "KHH", "KHH", "KHH", 
"KHH", "KHH", "KHH", "KHH", "KHH", "KHH", "KHH", "KHH", "KHH", 
"KHH", "KHH", "KHH", "YYR", "YYR", "YYR", "YYR", "YYR", "YYR", 
"YYR", "YYR"), n = c("Changzhi", "Chaochou", "Chaozhou", "Checheng", 
"Donggang", "Fangliao", "Fangshan", "Fengshan", "Gaoshu", "Hengchun Old Town", 
"Jiuru", "Linbian", "Liuqiu", "Manzhou", "Pingtung", "Tungkang", 
"Wanluan", "Yanpu", "Cartwright", "Happy Valley - Goose Bay", 
"Hopedale", "Makkovik", "Nain", "Natuashish", "Postville", "Rigolet"
), c = c("Taiwan", "Taiwan", "Taiwan", "Taiwan", "Taiwan", "Taiwan", 
"Taiwan", "Taiwan", "Taiwan", "Taiwan", "Taiwan", "Taiwan", "Taiwan", 
"Taiwan", "Taiwan", "Taiwan", "Taiwan", "Taiwan", "Canada", "Canada", 
"Canada", "Canada", "Canada", "Canada", "Canada", "Canada"), 
    C = c("TW", "TW", "TW", "TW", "TW", "TW", "TW", "TW", "TW", 
    "TW", "TW", "TW", "TW", "TW", "TW", "TW", "TW", "TW", "CA", 
    "CA", "CA", "CA", "CA", "CA", "CA", "CA"), r = c("Eastern Asia", 
    "Eastern Asia", "Eastern Asia", "Eastern Asia", "Eastern Asia", 
    "Eastern Asia", "Eastern Asia", "Eastern Asia", "Eastern Asia", 
    "Eastern Asia", "Eastern Asia", "Eastern Asia", "Eastern Asia", 
    "Eastern Asia", "Eastern Asia", "Eastern Asia", "Eastern Asia", 
    "Eastern Asia", "Northern America", "Northern America", "Northern America", 
    "Northern America", "Northern America", "Northern America", 
    "Northern America", "Northern America"), L = c(120.576, 120.539, 
    120.533, 120.717, 120.482, 120.593, 120.657, 120.35, 120.625, 
    120.744, 120.487, 120.488, 120.383, 120.848, 120.488, 120.449, 
    120.625, 120.574, -57.042, -60.377, -60.229, -59.186, -61.68, 
    -61.184, -59.785, -58.458), F = c(22.701, 22.55, 22.552, 
    22.06, 22.459, 22.368, 22.249, 22.633, 22.737, 22.005, 22.725, 
    22.428, 22.352, 21.991, 22.672, 22.465, 22.612, 22.76, 53.683, 
    53.31, 55.448, 55.077, 56.549, 55.914, 54.91, 54.18)), sorted = "i", class = c("data.table", 
"data.frame"), row.names = c(NA, -26L))

dput wasn't working with the list/data.table so I put it up as a dataframe. If it would help you to put it into a list then you can run split(df[, -1],df$i)

I would like to take the name of each row within each list and put it into a new list within each row.

I have a large list which looks like this:

    i                        n      c  C                r       L      F
1 YYR               Cartwright Canada CA Northern America -57.042 53.683
2 YYR Happy Valley - Goose Bay Canada CA Northern America -60.377 53.310
3 YYR                 Hopedale Canada CA Northern America -60.229 55.448
4 YYR                 Makkovik Canada CA Northern America -59.186 55.077
5 YYR                     Nain Canada CA Northern America -61.680 56.549   <-- this row is used in the example
6 YYR               Natuashish Canada CA Northern America -61.184 55.914
...

I'd like the output to look like this:

df$YYR

             n      c  C                r       L      F x
1:  Cartwright Canada CA Northern America -59.785 54.910 list()
...

df$YYR["n"][5,]

[1] "Nain"
...

df$YYR[["x"]][5] <-- this is the needed structure

[[5]]
                         d n      index
1                      YYR Nain   0
2               Cartwright Nain   1
3 Happy Valley - Goose Bay Nain   2
4                 Hopedale Nain   3
5                 Makkovik Nain   4
6               Natuashish Nain   6

...

toJSON(df$YYR["x"][5,]) <-- hopefully this makes sense because it is confusing me also

[
  {
    "d": "YYR",
    "n": "Nain",
    "index": 0
  },
  {
    "d": "Cartwright",
    "n": "Nain",
    "index": 1
  },
  {
    "d": "Happy Valley - Goose Bay",
    "n": "Nain",
    "index": 2
  },
  {
    "d": "Hopedale",
    "n": "Nain",
    "index": 3
  },
  {
    "d": "Makkovik",
    "n": "Nain",
    "index": 4
  },
...
]

similar to expand.grid or cartesian join within each list element but also including a reference to the parent list element as the 0 index value

Upvotes: 0

Views: 77

Answers (1)

eddi
eddi

Reputation: 49448

I don't fully understand what you want, but this will produce the example results:

res = df[, .(d = c(i, n), idx = 0:.N), by = i]
res[res[idx > 0], on = .(i), allow = T, .(d, n = i.d, idx)][d != n]
#              d        n idx
#  1:        KHH Changzhi   0
#  2:   Chaochou Changzhi   2
#  3:   Chaozhou Changzhi   3
#  4:   Checheng Changzhi   4
#  5:   Donggang Changzhi   5
# ---                        
#384:   Hopedale  Rigolet   3
#385:   Makkovik  Rigolet   4
#386:       Nain  Rigolet   5
#387: Natuashish  Rigolet   6
#388:  Postville  Rigolet   7

Upvotes: 1

Related Questions