通过从 R 中的列表中提取元素来创建向量

Question

我有一个列表对象，其组件如下所示：

$AA
(Intercept)   AA
3             4

$CD
(Intercept)   CD 
4             2

$XY
(Intercept)   XY 
5             3

我想要一个如下所示的数据框。 Alpha对应上面的(Intercept)，Beta对应列表项名称所代表的值。

Name    Alpha    Beta
AA      3        4
CD      4        2
XY      5        3

我知道我可以通过以下方式提取值，其中 coefs 是列表的名称。

coefs$AA[1]  #value for Alpha 
coefs$AA[2]  #value for Beta

如何对列表的所有组件执行此操作？如果有用，该列表表示几个回归输出的系数 alpha 和 beta。

数据：

dput(coefs)
list(AA = c(`(Intercept)` = 0.00781534742097672, AA = 0.140455432370463
), ALD = c(`(Intercept)` = 0.00557182631277353, ALD = 0.198916138245909
), AXP = c(`(Intercept)` = 0.00633440947245968, AXP = 0.154322769323201
), BA = c(`(Intercept)` = 0.00834280894501586, BA = 0.0941536953127183
), CAT = c(`(Intercept)` = 0.00655105648795371, CAT = 0.137462558039722
), CHV = c(`(Intercept)` = 0.0058599522637348, CHV = 0.269888033479484
), DD = c(`(Intercept)` = 0.00623849818068288, DD = 0.231206114050254
), DIS = c(`(Intercept)` = 0.0067614536257978, DIS = 0.180513015485431
), EK = c(`(Intercept)` = 0.00790519763382617, EK = 0.118639013243976
), ENE = c(`(Intercept)` = 0.00781778892551527, ENE = 0.10858254068385
), GE = c(`(Intercept)` = 0.00294196299492578, GE = 0.364461751377626
), GM = c(`(Intercept)` = 0.0085174289624691, GM = 0.0873909980420347
), GT = c(`(Intercept)` = 0.00707641628728448, GT = 0.204555410056181
), HWP = c(`(Intercept)` = 0.00634667846671781, HWP = 0.137844745901337
), IBM = c(`(Intercept)` = 0.00849044794002662, IBM = 0.0964252010051216
), IP = c(`(Intercept)` = 0.00906399011873481, IP = 0.151009935502848
), JNJ = c(`(Intercept)` = 0.00715601895187441, JNJ = 0.219180395262686
), JPM = c(`(Intercept)` = 0.00741476793645128, JPM = 0.241375375114064
), KO = c(`(Intercept)` = 0.00660478201497359, KO = 0.18143193126914
), MCD = c(`(Intercept)` = 0.00666717322231, MCD = 0.195659962104338
), MMM = c(`(Intercept)` = 0.00548410132538106, MMM = 0.331060750102674
), MO = c(`(Intercept)` = 0.00811094805656566, MO = 0.147779503536321
), MRK = c(`(Intercept)` = 0.00754935866336214, MRK = 0.237287608850202
), PG = c(`(Intercept)` = 0.00575628655569729, PG = 0.244866242495482
), S = c(`(Intercept)` = 0.00565228048065954, S = 0.175250230655151
), TRV = c(`(Intercept)` = 0.00392956375771792, TRV = 0.199917956865273
), UK = c(`(Intercept)` = 0.00627252118663475, UK = 0.113058380208163
), UTX = c(`(Intercept)` = 0.00417151421589462, UTX = 0.308220734271536
), WMT = c(`(Intercept)` = 0.0101499747347963, WMT = 0.1250276376077
), XON = c(`(Intercept)` = 0.0047869774616443, XON = 0.42015961399732
))

Answer 1

这是 do.call 的方法：

Result <- data.frame(Name = names(coefs),
                     do.call(rbind, coefs))
names(Result) <- c("Name","Alpha","Beta")
rownames(Result) <- NULL
Result
#   Name       Alpha      Beta
#1    AA 0.007815347 0.1404554
#2   ALD 0.005571826 0.1989161
#3   AXP 0.006334409 0.1543228
#4    BA 0.008342809 0.0941537
#5   CAT 0.006551056 0.1374626
#...

Answer 2

使用 purrr 的 map_df :

purrr::map_df(coefs, ~setNames(data.frame(t(.x)), c('alpha', 'beta')), .id = 'Name')

#   Name       alpha      beta
#1    AA 0.007815347 0.1404554
#2   ALD 0.005571826 0.1989161
#3   AXP 0.006334409 0.1543228
#4    BA 0.008342809 0.0941537
#5   CAT 0.006551056 0.1374626
#6   CHV 0.005859952 0.2698880
#...
#...

通过从 R 中的列表中提取元素来创建向量

Create vectors by extracting elements from list in R

r

extract

list

apply

tidyverse