How to create in R a list for a dataframe containg common character through an iterative function

Question

this is the dataframe with trice repeated comparisons for every outcome variable (so the submutiple level is 12)

 A tibble: 36 x 10
   s     .y.   group1          group2             n1    n2     p p.signif p.adj p.adj.signif
 * <chr> <chr> <chr>           <chr>           <int> <int> <dbl> <chr>    <dbl> <chr>       
 1 E     value new_value_for_8 new_value_for_6    25    25 0.877 ns       1     ns          
 2 E     value new_value_for_8 new_value_for_4    25    25 0.929 ns       1     ns          
 3 E     value new_value_for_6 new_value_for_4    25    25 0.948 ns       1     ns          
 4 F     value new_value_for_8 new_value_for_6    25    25 0.735 ns       1     ns          
 5 F     value new_value_for_8 new_value_for_4    25    25 0.738 ns       1     ns          
 6 F     value new_value_for_6 new_value_for_4    25    25 0.501 ns       1     ns          
 7 G     value new_value_for_8 new_value_for_6    25    25 0.808 ns       0.808 ns          
 8 G     value new_value_for_8 new_value_for_4    25    25 0.101 ns       0.303 ns          
 9 G     value new_value_for_6 new_value_for_4    25    25 0.161 ns       0.321 ns          
10 H     value new_value_for_8 new_value_for_6    25    25 0.964 ns       0.964 ns

I've just created a 12-element list for each of the three repetaed measures containing relative statistics as follows:

my_comparisons <- list(E = comparisons[1:3,], 
                       F = comparisons[4:6,], 
                       G = comparisons[7:9,], 
                       H = comparisons[10:12,], 
                       I = comparisons[13:15,], 
                       J = comparisons[16:18,], 
                       K =comparisons[19:21,], 
                       L = comparisons[22:24,], 
                       M = comparisons[25:27,], 
                       N = comparisons[28:30,], 
                       O  = comparisons[31:33,], 
                       P = comparisons[34:36,])

obtaining the following results

[[E]]

    # A tibble: 3 x 10
      s     .y.   group1          group2             n1    n2     p p.signif p.adj p.adj.signif
      <chr> <chr> <chr>           <chr>           <int> <int> <dbl> <chr>    <dbl> <chr>       
    1 E     value new_value_for_8 new_value_for_6    25    25 0.877 ns           1 ns          
    2 E     value new_value_for_8 new_value_for_4    25    25 0.929 ns           1 ns          
    3 E     value new_value_for_6 new_value_for_4    25    25 0.948 ns           1 ns  
    [[F]] .... and so on

Since how you could see in the signals columns, according to which I split the list there some common font (for instance P3, FCz, LPPearly and so on). I would like to use an iterate function such lapply(), some loops or map() functions in order to automatize this list creation.

Thanks in advance

Here the original dataset

    > dput(head(df, 10))
structure(list(A = 1:10, C = c("Maybe", "Maybe", "Maybe", "Maybe", 
"Maybe", "Maybe", "Maybe", "Maybe", "Maybe", "Maybe"), D = structure(c(1L, 
2L, 3L, 1L, 2L, 3L, 1L, 2L, 3L, 1L), .Label = c("new_value_for_8", 
"new_value_for_6", "new_value_for_4"), class = "factor"), E = c(988.368784828308, 
988.856158671407, 996.004085290553, 999.685844324618, 1000.23888564896, 
1005.03749946898, 999.786378084971, 997.039675082569, 998.028313183065, 
997.168905747014), F = c(994.834756009939, 994.468875098246, 
1000.62150212342, 1002.23100741241, 1003.96990710863, 1007.75899775608, 
998.699806256246, 996.401009591011, 998.076594704249, 1002.19344184533
), G = c(1011.88022669726, 1012.10534266625, 1012.9554415821, 
1015.09810043606, 1015.40462298842, 1016.67103699915, 1003.13771453335, 
999.9107434841, 1002.15365554737, 1013.67789244066), H = c(988.221495702721, 
990.850727928741, 992.418094914622, 995.984841639886, 993.398346143465, 
997.971380355398, 1004.4672957051, 1002.54036572775, 1002.2292388993, 
999.116379988893), I = c(994.035709684742, 994.890815628412, 
997.18267770374, 998.564426335124, 996.851278420874, 1000.16039368502, 
1003.52155765272, 1002.1043798945, 1002.7069399281, 1005.49897156208
), J = c(1008.23981597718, 1009.51261484649, 1009.42367409926, 
1005.06332653216, 1005.02619159395, 1009.07903916629, 1007.56089165218, 
1005.49719893791, 1004.91476855238, 1013.03209535721), K = c(994.327042030287, 
995.608170991922, 997.033470393412, 1000.15918365269, 998.216388150646, 
1001.97377908784, 1003.17401220482, 1001.60211665164, 1002.27932356239, 
1002.41479226363), L = c(999.225538268699, 999.349990537239, 
1001.14010250645, 1001.51403741206, 1000.25571835554, 1003.76051565494, 
1002.74763442988, 1001.09116707486, 1003.29833843754, 1006.55857216695
), M = c(1009.99385579756, 1011.12126521731, 1010.6989716872, 
1003.7899021821, 1004.59413830322, 1008.52123662618, 1006.34418311104, 
1004.1077131243, 1004.94124365003, 1011.89121961563), N = c(999.801263745036, 
996.838989582336, 1000.89599227983, 1003.11042068113, 1002.27800090558, 
1003.83846437952, 1000.70169995102, 1001.75290674649, 998.660833714301, 
1006.69246804854), O = c(1002.96437294923, 997.870867692911, 
1002.94619035116, 1003.44844607015, 1003.02403433836, 1004.70457675466, 
999.880559826981, 1000.66826545719, 999.59436981446, 1007.32640154038
), P = c(1006.28027312932, 1005.24535230967, 1007.68162285336, 
1001.08242973466, 1002.99896314, 1005.36085942954, 1001.22060069797, 
1000.43007709819, 1000.47666761108, 1008.73650967215)), row.names = c(NA, 
10L), class = "data.frame")

Have a look at [this answer](https://stackoverflow.com/questions/30944281/r-multi-index-on-columns-and-or-rows) — scandav, Oct 15 '21 at 12:20
Thansk. If you could to try replying with an extended answer below, run by using the dataset I've post it would be better. — 12666727b9, Oct 15 '21 at 14:44

PaulS · Accepted Answer · 2021-10-15T20:53:30.773

I guess, you are looking for something like the following:

library(dplyr)

df <- structure(list(signals = c("P3FCz", "P3FCz", "P3FCz", "P3Cz", 
"P3Cz", "P3Cz", "P3Pz", "P3Pz", "P3Pz", "LPPearlyFCz", "LPPearlyFCz", 
"LPPearlyFCz", "LPPearlyCz", "LPPearlyCz", "LPPearlyCz", "LPPearlyPz", 
"LPPearlyPz", "LPPearlyPz", "LPP1FCz", "LPP1FCz", "LPP1FCz", 
"LPP1Cz", "LPP1Cz", "LPP1Cz", "LPP1Pz", "LPP1Pz", "LPP1Pz", "LPP2FCz", 
"LPP2FCz", "LPP2FCz", "LPP2Cz", "LPP2Cz", "LPP2Cz", "LPP2Pz", 
"LPP2Pz", "LPP2Pz"), .y. = c("value", "value", "value", "value", 
"value", "value", "value", "value", "value", "value", "value", 
"value", "value", "value", "value", "value", "value", "value", 
"value", "value", "value", "value", "value", "value", "value", 
"value", "value", "value", "value", "value", "value", "value", 
"value", "value", "value", "value"), group1 = c("NEG-CTR", "NEG-CTR", 
"NEG-NOC", "NEG-CTR", "NEG-CTR", "NEG-NOC", "NEG-CTR", "NEG-CTR", 
"NEG-NOC", "NEG-CTR", "NEG-CTR", "NEG-NOC", "NEG-CTR", "NEG-CTR", 
"NEG-NOC", "NEG-CTR", "NEG-CTR", "NEG-NOC", "NEG-CTR", "NEG-CTR", 
"NEG-NOC", "NEG-CTR", "NEG-CTR", "NEG-NOC", "NEG-CTR", "NEG-CTR", 
"NEG-NOC", "NEG-CTR", "NEG-CTR", "NEG-NOC", "NEG-CTR", "NEG-CTR", 
"NEG-NOC", "NEG-CTR", "NEG-CTR", "NEG-NOC"), group2 = c("NEG-NOC", 
"NEU-NOC", "NEU-NOC", "NEG-NOC", "NEU-NOC", "NEU-NOC", "NEG-NOC", 
"NEU-NOC", "NEU-NOC", "NEG-NOC", "NEU-NOC", "NEU-NOC", "NEG-NOC", 
"NEU-NOC", "NEU-NOC", "NEG-NOC", "NEU-NOC", "NEU-NOC", "NEG-NOC", 
"NEU-NOC", "NEU-NOC", "NEG-NOC", "NEU-NOC", "NEU-NOC", "NEG-NOC", 
"NEU-NOC", "NEU-NOC", "NEG-NOC", "NEU-NOC", "NEU-NOC", "NEG-NOC", 
"NEU-NOC", "NEU-NOC", "NEG-NOC", "NEU-NOC", "NEU-NOC"), n1 = c(25L, 
25L, 25L, 25L, 25L, 25L, 25L, 25L, 25L, 25L, 25L, 25L, 25L, 25L, 
25L, 25L, 25L, 25L, 25L, 25L, 25L, 25L, 25L, 25L, 25L, 25L, 25L, 
25L, 25L, 25L, 25L, 25L, 25L, 25L, 25L, 25L), n2 = c(25L, 25L, 
25L, 25L, 25L, 25L, 25L, 25L, 25L, 25L, 25L, 25L, 25L, 25L, 25L, 
25L, 25L, 25L, 25L, 25L, 25L, 25L, 25L, 25L, 25L, 25L, 25L, 25L, 
25L, 25L, 25L, 25L, 25L, 25L, 25L, 25L), statistic = c(-0.32183284, 
-0.17788461, 0.11249149, -0.62380748, 0.59236111, 0.92477314, 
0.43979736, 3.10746654, 2.4289231, -0.09188784, 2.31385915, 2.30243506, 
-0.36897352, 3.28159273, 3.09240265, 0.06703844, 4.2591323, 4.43158703, 
-0.39439158, 2.53611856, 2.36271993, -1.06592362, 2.77405996, 
3.06325458, -0.54210261, 3.72755117, 4.31056245, -0.58228303, 
0.10238271, 0.58654953, -1.32163941, 0.02393817, 1.13763114, 
-1.63511147, 0.8700396, 2.10635863), df = c(24L, 24L, 24L, 24L, 
24L, 24L, 24L, 24L, 24L, 24L, 24L, 24L, 24L, 24L, 24L, 24L, 24L, 
24L, 24L, 24L, 24L, 24L, 24L, 24L, 24L, 24L, 24L, 24L, 24L, 24L, 
24L, 24L, 24L, 24L, 24L, 24L), p = c(0.75, 0.86, 0.911, 0.539, 
0.559, 0.364, 0.664, 0.005, 0.023, 0.928, 0.03, 0.03, 0.715, 
0.003, 0.005, 0.947, 0.000273, 0.000176, 0.697, 0.018, 0.027, 
0.297, 0.011, 0.005, 0.593, 0.001, 0.00024, 0.566, 0.919, 0.563, 
0.199, 0.981, 0.267, 0.115, 0.393, 0.046), p.adj = c(1, 1, 1, 
1, 1, 1, 1, 0.014, 0.069, 1, 0.089, 0.091, 1, 0.009, 0.015, 1, 
0.000819, 0.000528, 1, 0.054, 0.08, 0.891, 0.032, 0.016, 1, 0.003, 
0.00072, 1, 1, 1, 0.597, 1, 0.801, 0.345, 1, 0.137), p.adj.signif = c("ns", 
"ns", "ns", "ns", "ns", "ns", "ns", "*", "ns", "ns", "ns", "ns", 
"ns", "**", "*", "ns", "***", "***", "ns", "ns", "ns", "ns", 
"*", "*", "ns", "**", "***", "ns", "ns", "ns", "ns", "ns", "ns", 
"ns", "ns", "\n")), row.names = c(NA, -36L), class = "data.frame")

df %>% 
  group_split(signals) %>% 
  as.list() %>% 
  setNames(sort(unique(df$signals)))

Exactly. Thanks....and what if I would like to rename each object in it with each 'signals' name? — 12666727b9, Oct 15 '21 at 20:49
I have just edited my code by adding a last line: with that additional line, you get what you want. — PaulS, Oct 15 '21 at 20:54
Many thanks man. You've helped to fix problem. I've turned this question since I think it could me help to make me out from another problem I'm in trouble with.....it connected to get label from this list (and from the anova test list as well as I explained here https://stackoverflow.com/questions/69528926/how-to-print-on-a-serie-sof-graphs-pairwise-comparisons-bars-and-effect-size-val). Please feel free to give your contribute if you are willing to ..Thanks again — 12666727b9, Oct 15 '21 at 21:11

How to create in R a list for a dataframe containg common character through an iterative function

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