I think I need to restate the problem. If the test data is only a vector,
then I am predicting one test sample. But the output from the predict result
has the same length as the training set. And there is a warning message
about this.

Annie

On Sat, Aug 8, 2009 at 2:52 PM, annie Zhang <annie.zhang2...@gmail.com>wrote:

> Hi, Milton,
>
> Thank you for the reply. I tried, but it seems the problem is the column
> name of the test data is not the same as the column name of the training
> data. I didn't give the column name, the system seemed do. How to chang
> here?
>
> Annie
>
>  On Fri, Aug 7, 2009 at 7:52 PM, milton ruser <milton.ru...@gmail.com>wrote:
>
>> Hi Annie,
>>
>> create a new data.frame with input variables having all predictors
>> variables on it.
>> after give a look at ?predict
>>
>> best wishes
>>
>> milton
>>
>>   On Fri, Aug 7, 2009 at 8:19 PM, annie Zhang 
>> <annie.zhang2...@gmail.com>wrote:
>>
>>>  Hi, R users,
>>>
>>> I am trying to use glm to do logistic regression. I know generally when I
>>> have two covariates, say x1 and x2, then I do
>>> fit <- glm(y~x1+x2,famliy='binomial')
>>> But now my covariates form a n*p matrix, say x, so actually each column
>>> is a
>>> covariate. So I think I should do
>>> fit <- glm(y~x,family='binomial')
>>> Then I need to predict new data. How should I write the newdata? I tried
>>> several thing, all failed. The x in the fit is a matrix, but is a vector
>>> for
>>> the new data.
>>>
>>> Thank you,
>>> Annie
>>>
>>>        [[alternative HTML version deleted]]
>>>
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>>> PLEASE do read the posting guide
>>> http://www.R-project.org/posting-guide.html<http://www.r-project.org/posting-guide.html>
>>> and provide commented, minimal, self-contained, reproducible code.
>>>
>>
>>
>

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