如何在 R 中通过迭代函数为包含公共字符的数据框创建列表 [英] How to create in R a list for a dataframe containg common character through an iterative function
本文介绍了如何在 R 中通过迭代函数为包含公共字符的数据框创建列表的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
这是对每个结果变量进行三次重复比较的数据框(因此子多级为 12)
this is the dataframe with trice repeated comparisons for every outcome variable (so the submutiple level is 12)
> as.data.frame(comparisons)
signals .y. group1 group2 n1 n2 statistic df p p.adj p.adj.signif
1 P3FCz value NEG-CTR NEG-NOC 25 25 -0.32183284 24 0.750000 1.000000 ns
2 P3FCz value NEG-CTR NEU-NOC 25 25 -0.17788461 24 0.860000 1.000000 ns
3 P3FCz value NEG-NOC NEU-NOC 25 25 0.11249149 24 0.911000 1.000000 ns
4 P3Cz value NEG-CTR NEG-NOC 25 25 -0.62380748 24 0.539000 1.000000 ns
5 P3Cz value NEG-CTR NEU-NOC 25 25 0.59236111 24 0.559000 1.000000 ns
6 P3Cz value NEG-NOC NEU-NOC 25 25 0.92477314 24 0.364000 1.000000 ns
7 P3Pz value NEG-CTR NEG-NOC 25 25 0.43979736 24 0.664000 1.000000 ns
8 P3Pz value NEG-CTR NEU-NOC 25 25 3.10746654 24 0.005000 0.014000 *
9 P3Pz value NEG-NOC NEU-NOC 25 25 2.42892310 24 0.023000 0.069000 ns
10 LPPearlyFCz value NEG-CTR NEG-NOC 25 25 -0.09188784 24 0.928000 1.000000 ns
11 LPPearlyFCz value NEG-CTR NEU-NOC 25 25 2.31385915 24 0.030000 0.089000 ns
12 LPPearlyFCz value NEG-NOC NEU-NOC 25 25 2.30243506 24 0.030000 0.091000 ns
13 LPPearlyCz value NEG-CTR NEG-NOC 25 25 -0.36897352 24 0.715000 1.000000 ns
14 LPPearlyCz value NEG-CTR NEU-NOC 25 25 3.28159273 24 0.003000 0.009000 **
15 LPPearlyCz value NEG-NOC NEU-NOC 25 25 3.09240265 24 0.005000 0.015000 *
16 LPPearlyPz value NEG-CTR NEG-NOC 25 25 0.06703844 24 0.947000 1.000000 ns
17 LPPearlyPz value NEG-CTR NEU-NOC 25 25 4.25913230 24 0.000273 0.000819 ***
18 LPPearlyPz value NEG-NOC NEU-NOC 25 25 4.43158703 24 0.000176 0.000528 ***
19 LPP1FCz value NEG-CTR NEG-NOC 25 25 -0.39439158 24 0.697000 1.000000 ns
20 LPP1FCz value NEG-CTR NEU-NOC 25 25 2.53611856 24 0.018000 0.054000 ns
21 LPP1FCz value NEG-NOC NEU-NOC 25 25 2.36271993 24 0.027000 0.080000 ns
22 LPP1Cz value NEG-CTR NEG-NOC 25 25 -1.06592362 24 0.297000 0.891000 ns
23 LPP1Cz value NEG-CTR NEU-NOC 25 25 2.77405996 24 0.011000 0.032000 *
24 LPP1Cz value NEG-NOC NEU-NOC 25 25 3.06325458 24 0.005000 0.016000 *
25 LPP1Pz value NEG-CTR NEG-NOC 25 25 -0.54210261 24 0.593000 1.000000 ns
26 LPP1Pz value NEG-CTR NEU-NOC 25 25 3.72755117 24 0.001000 0.003000 **
27 LPP1Pz value NEG-NOC NEU-NOC 25 25 4.31056245 24 0.000240 0.000720 ***
28 LPP2FCz value NEG-CTR NEG-NOC 25 25 -0.58228303 24 0.566000 1.000000 ns
29 LPP2FCz value NEG-CTR NEU-NOC 25 25 0.10238271 24 0.919000 1.000000 ns
30 LPP2FCz value NEG-NOC NEU-NOC 25 25 0.58654953 24 0.563000 1.000000 ns
31 LPP2Cz value NEG-CTR NEG-NOC 25 25 -1.32163941 24 0.199000 0.597000 ns
32 LPP2Cz value NEG-CTR NEU-NOC 25 25 0.02393817 24 0.981000 1.000000 ns
33 LPP2Cz value NEG-NOC NEU-NOC 25 25 1.13763114 24 0.267000 0.801000 ns
34 LPP2Pz value NEG-CTR NEG-NOC 25 25 -1.63511147 24 0.115000 0.345000 ns
35 LPP2Pz value NEG-CTR NEU-NOC 25 25 0.87003960 24 0.393000 1.000000 ns
36 LPP2Pz value NEG-NOC NEU-NOC 25 25 2.10635863 24 0.046000 0.137000 ns
>
我刚刚为三个重复测量的每一个创建了一个 12 元素列表，其中包含如下相关统计数据:
I've just created a 12-element list for each of the three repetaed measures containing relative statistics as follows:
my_comparisons <- list(P3FCz = comparisons[1:3,],
P3Cz = comparisons[4:6,],
P3Pz = comparisons[7:9,],
LPPearlyFcz = comparisons[10:12,],
LPPearlyCz = comparisons[13:15,],
LPPearlyPz = comparisons[16:18,],
LPP1FCz =comparisons[19:21,],
LPP1Cz = comparisons[22:24,],
LPP1Pz = comparisons[25:27,],
LPP2FCz = comparisons[28:30,],
LPP2Cz = comparisons[31:33,],
LPP2Pz = comparisons[34:36,])
得到以下结果
[[P3FCz]]
y. group1 group2 n1 n2 statistic df p.......
value NEG-CTR NEG-NOC 25 25 -0.32183284 24 0.750000 ...
value NEG-CTR NEU-NOC 25 25 -0.17788461 24 0.860000 ....
value NEG-NOC NEU-NOC 25 25 0.11249149 24 0.911000 ....
[[P3Cz]] .... and so on
由于您可以在信号列中看到，因此我将列表拆分为一些常见字体(例如 P3、FCz、LPPearly 等).我想使用迭代函数，例如 lapply()、一些循环或 map() 函数，以便自动创建此列表.
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.
提前致谢
这里是原始数据集
> dput(head(df_join))
structure(list(ID = c("01", "01", "01", "04", "04", "04"), GR = c("RP",
"RP", "RP", "RP", "RP", "RP"), SES = c("V", "V", "V", "V", "V",
"V"), COND = structure(c(1L, 2L, 3L, 1L, 2L, 3L), .Label = c("NEG-CTR",
"NEG-NOC", "NEU-NOC"), class = "factor"), P3FCz = c(-11.6312151716924,
-11.1438413285935, -3.99591470944713, -0.314155675382471, 0.238885648959708,
5.03749946898385), P3Cz = c(-5.16524399006139, -5.53112490175437,
0.621502123415388, 2.23100741241039, 3.96990710862955, 7.75899775608441
), P3Pz = c(11.8802266972569, 12.1053426662461, 12.955441582096,
15.0981004360619, 15.4046229884164, 16.671036999147), LPPearlyFCz = c(-11.7785042972793,
-9.14927207125904, -7.58190508537766, -4.01515836011381, -6.60165385653499,
-2.02861964460179), LPPearlyCz = c(-5.96429031525769, -5.10918437158799,
-2.81732229625975, -1.43557366487622, -3.14872157912645, 0.160393685024631
), LPPearlyPz = c(8.23981597718437, 9.51261484648731, 9.42367409925817,
5.06332653216481, 5.02619159395405, 9.07903916629231), LPP1FCz = c(-5.67295796971287,
-4.3918290080777, -2.96652960658775, 0.159183652691071, -1.78361184935376,
1.97377908783621), LPP1Cz = c(-0.774461731301161, -0.650009462761383,
1.14010250644923, 1.51403741206392, 0.25571835554024, 3.76051565494304
), LPP1Pz = c(9.99385579756163, 11.1212652173052, 10.6989716871958,
3.7899021820967, 4.59413830322224, 8.52123662617732), LPP2FCz = c(-0.198736254963744,
-3.16101041766438, 0.895992279831378, 3.11042068112836, 2.27800090558473,
3.83846437952292), LPP2Cz = c(2.96437294922766, -2.12913230708907,
2.94619035115619, 3.44844607014521, 3.02403433835637, 4.7045767546583
), LPP2Pz = c(6.28027312932027, 5.24535230966772, 7.68162285335806,
1.08242973465635, 2.99896314000211, 5.36085942954182)), row.names = c(NA,
6L), class = "data.frame")
>
推荐答案
我猜，你正在寻找类似下面的东西:
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", "
")), row.names = c(NA, -36L), class = "data.frame")
df %>%
group_split(signals) %>%
as.list() %>%
setNames(sort(unique(df$signals)))
这篇关于如何在 R 中通过迭代函数为包含公共字符的数据框创建列表的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
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