r - multinom() 默认如何处理 NA 值？

Question

当我运行时multinom()，比如说Y ~ X1 + X2 + X3，如果对于一个特定的行X1是NA（即缺失），但是Y，X2并且X3都有一个值，那么整行是否会被丢弃（就像在 SAS 中一样）？中的缺失值如何处理multinom()？

Answer 1

这是一个简单的例子（来自?multinom包nnet）来探索不同的na.action：

> library(nnet)
> library(MASS)
> example(birthwt)
> (bwt.mu <- multinom(low ~ ., bwt))

有意创造NA价值：

> bwt[1,"age"]<-NA # Intentionally create NA value
> nrow(bwt)
[1] 189

测试4种不同的na.action：

> predict(multinom(low ~ ., bwt, na.action=na.exclude)) # Note length is 189
# weights:  12 (11 variable)
initial  value 130.311670
iter  10 value 97.622035
final  value 97.359978
converged
  [1] <NA> 0    0    0    0    0    0    0    0    0    0    0    1    1    0
 [16] 0    0    0    0    0    0    0    0    0    0    0    0    0    0    0
 ....

> predict(multinom(low ~ ., bwt, na.action=na.omit)) # Note length is 188
# weights:  12 (11 variable)
initial  value 130.311670
iter  10 value 97.622035
final  value 97.359978
converged
  [1] 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0
 [38] 0 0 0 0 0 0 1 1 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 1 1 0 0 0 0
 .....

> predict(multinom(low ~ ., bwt, na.action=na.fail))    # Generates error
Error in na.fail.default(list(low = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,  :
  missing values in object

> predict(multinom(low ~ ., bwt, na.action=na.pass))    # Generates error
Error in qr.default(X) : NA/NaN/Inf in foreign function call (arg 1)

因此在预测中na.exclude生成 a而完全省略它。并且不会创建模型。如果未指定，则显示默认值：NAna.omitna.passna.failna.action

> getOption("na.action")
[1] "na.omit"

Answer 2

您可以指定行为

- na.omit and na.exclude: returns the object with observations removed if they contain any missing values; differences between omitting and excluding NAs can be seen in some prediction and residual functions
- na.pass: returns the object unchanged
- na.fail: returns the object only if it contains no missing values

http://www.ats.ucla.edu/stat/r/faq/missing.htm