我有一个包含多个数据框的列表 -cj1-
dput(head(cj1[1:2]))
list(structure(list(individual = c("a12TTT.pdf", "a15.pdf", "a17.pdf",
"a18.pdf", "a21.pdf", "a2TTT.pdf", "a5.pdf", "B11.pdf", "B12.pdf",
"B13.pdf", "B22.pdf", "B24.pdf", "B4.pdf", "B7.pdf", "B8.pdf",
"cw10-1.pdf", "cw13-1.pdf", "cw15-1TTT.pdf", "cw17-1.pdf", "cw18.pdf",
"cw3.pdf", "cw4.pdf", "cw7_1TTT.pdf"), id = 1:23, Ntot = c(13,
9, 16, 15, 9, 13, 10, 10, 11, 10, 14, 10, 11, 12, 11, 10, 15,
12, 14, 11, 9, 10, 11), N1 = c(5, 5, 10, 11, 7, 9, 5, 5, 6, 8,
8, 8, 9, 8, 7, 1, 0, 6, 3, 4, 2, 4, 2), ND = c(0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0), N0 = c(8,
4, 6, 4, 2, 4, 5, 5, 5, 2, 6, 2, 2, 4, 4, 9, 15, 6, 11, 7, 7,
6, 9), score = c(5.06923076923077, 4.96666666666667, 9.925, 10.86,
6.83333333333333, 8.88461538461539, 5, 5, 5.97272727272727, 7.82,
7.95714285714286, 7.82, 8.80909090909091, 7.9, 6.91818181818182,
1.24, 0.3, 6, 3.17142857142857, 4.08181818181818, 2.16666666666667,
4.06, 2.19090909090909), propscore = c(0.389940828402367, 0.551851851851852,
0.6203125, 0.724, 0.759259259259259, 0.683431952662722, 0.5,
0.5, 0.54297520661157, 0.782, 0.568367346938776, 0.782, 0.800826446280992,
0.658333333333333, 0.628925619834711, 0.124, 0.02, 0.5, 0.226530612244898,
0.371074380165289, 0.240740740740741, 0.406, 0.199173553719008
), theta = c(-0.571211122446447, 0.418736780198501, 0.464533662219296,
0.760432013134893, 1.43961032059382, 0.935963883364303, 0.0742361005467161,
0.416783201347136, 0.232586422933618, 1.65345248955369, 0.178947462869717,
1.3980442736112, 1.5300599487058, 0.340087410746963, 0.616985944469495,
-1.73246102772711, -4.06186172096556, -0.347700710331151, -1.21009964741398,
0.239145600406579, -1.88836418690337, -0.276451472526056, -0.611455626388059
), se.theta = c(0.689550115014498, 0.689441554709003, 0.595659709892116,
0.609506508256404, 0.917792293663691, 0.652011367164736, 0.720534163064516,
0.695969555549033, 0.661019531367007, 0.87050969318314, 0.605775647419845,
0.797443937820774, 0.768436114096332, 0.695748274310803, 0.709380679025605,
1.00089414765463, 1.8701468050665, 0.68959824350285, 0.733014089189809,
0.656392513303483, 0.952935324276941, 0.71608982789968, 0.771906532861938
), outfit = c(1.24922700170817, 1.46067763769417, 0.915183304626819,
0.753992664091072, 0.37410361433915, 0.727316037037668, 0.616907868814702,
1.01528298230254, 1.01594232662062, 0.616808170683195, 0.646097057961938,
0.622993494551005, 0.807441271101246, 0.788526018181888, 1.2157399735092,
0.341189086206191, 0.021052091633073, 0.543024513106335, 1.04183076617928,
1.1772656963046, 0.736106160865241, 0.756316095787985, 0.58320701094964
), infit = c(1.4078580948461, 1.42854494963967, 1.09762978932861,
0.893957122448352, 0.64936943769433, 0.899191443180872, 0.724956556509282,
1.14975990693782, 1.08074439712469, 0.978248081241133, 0.755557633771936,
0.823903684368671, 0.911855771375284, 0.954272320131035, 0.926253596526142,
0.634052701587448, 0.0504659822408584, 0.712539957033173, 0.966034039620798,
1.1901663169553, 0.81371119642719, 0.817417869881874, 0.737574872116582
)), row.names = c(NA, -23L), class = "data.frame"), structure(list(
parlabel = c("Ties", "Home"), par = c("delta", "eta"), est = c(-43.5016417611571,
0.337872999554289), se = c(366043197.615422, 0.215169736220537
)), row.names = c(NA, -2L), class = "data.frame"))
head(cj1[[1]],2)
individual id Ntot N1 ND N0 score propscore theta se.theta outfit
1 a12TTT.pdf 1 13 5 0 8 5.069231 0.3899408 -0.5712111 0.6895501 1.249227
2 a15.pdf 2 9 5 0 4 4.966667 0.5518519 0.4187368 0.6894416 1.460678
infit
1 1.407858
2 1.428545
individual 和 theta 的第 1 列和第 9 列的数据框我试过:
results1<-sapply(cj1, "[",c("individual","theta") )
[.data.frame (X[[i]], ...) 中的错误:选择了未定义的列library(dplyr)
> results1 <- lapply(cj1, function(x) x%>% select(individual,theta))
Can't subset columns that don't exist.
x Columnindividualdoesn't exist.
Runrlang::last_error()to see where the error occurred.
我可以从一个数据框中减去这些列:
cj[[1]][c(1,9)]
最佳答案
您可以使用以下解决方案。我们使用 .x 来引用列表中的每个单独元素。这里 .x 可以是您的每个数据帧，我们只想选择其中的 2 列 c("individual","theta") 。
但是，由于只有一个数据框包含这样的列名，所以我使用 keep 函数实际上只保留其数据框包含所需列名的元素。请记住，对于这种称为 purrr-style 公式的编码形式，我们需要在 ~ 之前使用 .x 。因此，您使用 map 函数，它与基础 R 中的 lapply 等效，并使用此语法将任何函数应用于每个单独的元素(此处为数据帧)。
library(purrr)
cj1 %>%
map_if(~ all(c("individual","theta") %in% names(.x)),
~ .x %>% select(individual, theta)) %>%
keep(~ all(c("individual","theta") %in% names(.x)))
[[1]]
individual theta
1 a12TTT.pdf -0.5712111
2 a15.pdf 0.4187368
3 a17.pdf 0.4645337
4 a18.pdf 0.7604320
5 a21.pdf 1.4396103
6 a2TTT.pdf 0.9359639
7 a5.pdf 0.0742361
8 B11.pdf 0.4167832
9 B12.pdf 0.2325864
10 B13.pdf 1.6534525
11 B22.pdf 0.1789475
12 B24.pdf 1.3980443
13 B4.pdf 1.5300599
14 B7.pdf 0.3400874
15 B8.pdf 0.6169859
16 cw10-1.pdf -1.7324610
17 cw13-1.pdf -4.0618617
18 cw15-1TTT.pdf -0.3477007
19 cw17-1.pdf -1.2100996
20 cw18.pdf 0.2391456
21 cw3.pdf -1.8883642
22 cw4.pdf -0.2764515
23 cw7_1TTT.pdf -0.6114556
cj1 %>%
keep(~ all(c("individual","theta") %in% names(.x))) %>%
map(~ .x %>% select(individual, theta))
[[1]]
individual theta
1 a12TTT.pdf -0.5712111
2 a15.pdf 0.4187368
3 a17.pdf 0.4645337
4 a18.pdf 0.7604320
5 a21.pdf 1.4396103
6 a2TTT.pdf 0.9359639
7 a5.pdf 0.0742361
8 B11.pdf 0.4167832
9 B12.pdf 0.2325864
10 B13.pdf 1.6534525
11 B22.pdf 0.1789475
12 B24.pdf 1.3980443
13 B4.pdf 1.5300599
14 B7.pdf 0.3400874
15 B8.pdf 0.6169859
16 cw10-1.pdf -1.7324610
17 cw13-1.pdf -4.0618617
18 cw15-1TTT.pdf -0.3477007
19 cw17-1.pdf -1.2100996
20 cw18.pdf 0.2391456
21 cw3.pdf -1.8883642
22 cw4.pdf -0.2764515
23 cw7_1TTT.pdf -0.6114556
\(x) 等效于 function(x) ，这是 R. 4.1.0 可用的新功能。cj1 |>
lapply(\(x) {
if(all(c("individual","theta") %in% names(x))) {
`[`(x, c("individual","theta"))
}
}
) -> cj2
cj2 <- cj2[-which(sapply(cj2, is.null))] |> as.data.frame()
关于r - 从单独的数据框列表中的列表中的数据框中提取列，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/67858480/