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私はcbind lm summary別のリストに保存しようとしています。cbind20種類ほどのモデルが保管されているので、手ぶらで避けたいと思います。

どうすればこれができるか分かりますか?

ここに私のサンプルがありますlist

data = list(structure(list(`a alone` = structure(c(4L, 5L, 1L, 2L, 3L
), .Names = c("(Intercept)", "isexMALE", "numchild_rec1", "numchild_rec2", 
"numchild_rec>2"), .Label = c("-59.819", "-61.193", "-72.312", 
"126.679", "9.825"), class = "factor"), pvalue = structure(c(2L, 
1L, 2L, 2L, 2L), .Names = c("(Intercept)", "isexMALE", "numchild_rec1", 
"numchild_rec2", "numchild_rec>2"), .Label = c(" ", "***"), class =         "factor")), .Names = c("a alone", 
"pvalue"), row.names = c("(Intercept)", "isexMALE", "numchild_rec1", 
"numchild_rec2", "numchild_rec>2"), class = "data.frame"), structure(list(
    `b partner` = structure(c(4L, 5L, 1L, 2L, 3L), .Names = c("(Intercept)", 
"isexMALE", "numchild_rec1", "numchild_rec2", "numchild_rec>2"
), .Label = c("-222.064", "-259.233", "-277.213", "365.149", 
"8.608"), class = "factor"), pvalue = structure(c(2L, 1L, 
2L, 2L, 2L), .Names = c("(Intercept)", "isexMALE", "numchild_rec1", 
"numchild_rec2", "numchild_rec>2"), .Label = c(" ", "***"
), class = "factor")), .Names = c("b partner", "pvalue"), row.names = c("(Intercept)", 
"isexMALE", "numchild_rec1", "numchild_rec2", "numchild_rec>2"
), class = "data.frame"), structure(list(`c child` = structure(c(4L, 
1L, 5L, 2L, 3L), .Names = c("(Intercept)", "isexMALE", "numchild_rec1", 
"numchild_rec2", "numchild_rec>2"), .Label = c("-36.267", "112.03", 
"119.228", "23.706", "86.989"), class = "factor"), pvalue = structure(c(1L, 
2L, 2L, 2L, 2L), .Names = c("(Intercept)", "isexMALE", "numchild_rec1", 
"numchild_rec2", "numchild_rec>2"), .Label = c("**", "***"), class =     "factor")), .Names = c("c child", 
"pvalue"), row.names = c("(Intercept)", "isexMALE", "numchild_rec1", 
 "numchild_rec2", "numchild_rec>2"), class = "data.frame"))

このような出力を得たい

               a alone pvalue b partner pvalue c child pvalue
(Intercept)    126.679    ***   365.149    ***  23.706     **
isexMALE         9.825            8.608        -36.267    ***
numchild_rec1  -59.819    ***  -222.064    ***  86.989    ***
numchild_rec2  -61.193    ***  -259.233    ***  112.03    ***
numchild_rec>2 -72.312    ***  -277.213    *** 119.228    ***
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2 に答える 2

4

これを試して:

 data<- as.data.frame(data)
于 2015-07-01T11:55:20.253 に答える