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我有一个名为“mylist”的列表，其中包含 gam 输出。第一个列表的摘要如下：

> summary(mylist[[1]][[1]])

Family: quasipoisson 
Link function: log 

Formula:
cardva ~ s(trend, k = 11 * 6, fx = T, bs = "cr") + s(temp_01,      k = 6, fx = F, bs = "cr") + rh_01 + as.factor(dow) + s(fluepi,     k = 4, fx = F, bs = "cr") + as.factor(holiday) + Lag(pm1010,    0)

Parametric coefficients:
                      Estimate Std. Error t value Pr(>|t|)    
(Intercept)          3.1584139  0.0331388  95.309  < 2e-16 ***
rh_01                0.0005441  0.0004024   1.352  0.17639    
as.factor(dow)2      0.0356757  0.0127979   2.788  0.00533 ** 
as.factor(dow)3      0.0388823  0.0128057   3.036  0.00241 ** 
as.factor(dow)4      0.0107302  0.0129014   0.832  0.40561    
as.factor(dow)5      0.0243382  0.0128705   1.891  0.05867 .  
as.factor(dow)6      0.0277954  0.0128360   2.165  0.03040 *  
as.factor(dow)7      0.0275593  0.0127373   2.164  0.03053 *  
as.factor(holiday)1  0.0444349  0.0147219   3.018  0.00255 ** 
Lag(pm1010, 0)      -0.0010816  0.0042891  -0.252  0.80091

取消列表后，我提取了第一个列表的线性项的系数：

> head(plist)
[[1]]
                         Estimate   Std. Error    t value    Pr(>|t|)
(Intercept)          3.1584139271 0.0331388386 95.3085280 0.000000000
rh_01                0.0005441175 0.0004024202  1.3521128 0.176392590
as.factor(dow)2      0.0356757100 0.0127979429  2.7876128 0.005327293
as.factor(dow)3      0.0388823055 0.0128056733  3.0363343 0.002405504
as.factor(dow)4      0.0107302325 0.0129013816  0.8317119 0.405606249
as.factor(dow)5      0.0243382447 0.0128704711  1.8910143 0.058672841
as.factor(dow)6      0.0277953708 0.0128359850  2.1654256 0.030396240
as.factor(dow)7      0.0275592574 0.0127372874  2.1636677 0.030531063
as.factor(holiday)1  0.0444348611 0.0147218816  3.0182868 0.002553265
Lag(pm1010, 0)      -0.0010816252 0.0042890866 -0.2521808 0.800910389

我的问题是：是否可以将因变量的名称（在本例中为心脏）作为 plist 的一部分？

我想要达到的是（故意减少输出）

cardva          Estimate   Std. Error    t value    Pr(>|t|)
(Intercept)          3.1584139271 0.0331388386 95.3085280 0.000000000
rh_01                0.0005441175 0.0004024202  1.3521128 0.176392590
as.factor(dow)2      0.0356757100 0.0127979429  2.7876128 0.005327293

或者

                        Estimate   Std. Error    t value    Pr(>|t|) 
(Intercept)          3.1584139271 0.0331388386 95.3085280 0.000000000
rh_01                0.0005441175 0.0004024202  1.3521128 0.176392590
as.factor(dow)7      0.0275592574 0.0127372874  2.1636677 0.030531063
as.factor(holiday)1  0.0444348611 0.0147218816  3.0182868 0.002553265
cardva_Lag(pm1010, 0)      -0.0010816252 0.0042890866 -0.2521808 0.800910389
4

1 回答 1

1

两个选项： 命名列表的节点，然后将它们打印为：

names(plist)[1] <- 'cardva'
plist[1]
$cardva
                         Estimate   Std. Error    t value    Pr(>|t|)
(Intercept)          3.1584139271 0.0331388386 95.3085280 0.000000000
rh_01                0.0005441175 0.0004024202  1.3521128 0.176392590
as.factor(dow)2      0.0356757100 0.0127979429  2.7876128 0.005327293
as.factor(dow)3      0.0388823055 0.0128056733  3.0363343 0.002405504
as.factor(dow)4      0.0107302325 0.0129013816  0.8317119 0.405606249
as.factor(dow)5      0.0243382447 0.0128704711  1.8910143 0.058672841
as.factor(dow)6      0.0277953708 0.0128359850  2.1654256 0.030396240
as.factor(dow)7      0.0275592574 0.0127372874  2.1636677 0.030531063
as.factor(holiday)1  0.0444348611 0.0147218816  3.0182868 0.002553265
Lag(pm1010, 0)      -0.0010816252 0.0042890866 -0.2521808 0.800910389

或者：

temp <- plist[[1]] 
rownames(temp)[nrow(temp)] <- paste0( "cardva_", rownames(temp)[nrow(temp)]  )
于 2013-11-13T23:27:35.057 回答