<p>Unlike <a href="https://en.wikipedia.org/wiki/Proportional_hazards_model" rel="nofollow noreferrer">wikipedia</a>, the linear predictor (LP) in <code>coxph</code> is made dependent from the baseline hazard function <span class="math-container">$\lambda$</span>. For example, if a model has two predictors <code>x1</code>, <code>x2</code> stratified by <code>sex</code>, the linear predictor would be like:
<span class="math-container">$$LP = log (  \lambda _{sex}) + \beta _{1}x _{1}+\beta _{2}x _{2}$$</span></p>
<p>Where the <span class="math-container">$log(\lambda _{sex})$</span> can be obtained by <code>predict()</code> with <code>x1=x2=0</code>.</p>
<p>We could calculate the LP without strata information (sex) by hand using the <span class="math-container">$\beta _{1} \, \beta _{2}$</span> provided by <code>coef(fit)</code> and a customised <span class="math-container">$\lambda _{unkown \, sex}$</span>.</p>
<p>Below is a simulation comparing the results from <code>predict()</code> and by hand <code>coef(fit)</code>.</p>
<pre><code>## Toy data
df &lt;- list(time=c(4,3,1,1,2,2,3), 
          status=c(1,1,1,0,1,1,0), 
          x1=c(0,2,1,1,1,0,0), 
          x2=1:7,
          sex=c(0,0,0,0,1,1,1)) 

## Fit a stratified model 
fit &lt;- coxph(Surv(time, status) ~ x1 + x2 + strata(sex), df) 
coef(fit)
#         x1          x2 
# 0.79451881 -0.01917633

## Baseline linear predictor
lambda_sex1 &lt;- predict(fit, newdata=list(x1=0, x2=0, sex=0))
# -0.746578
lambda_sex2 &lt;- predict(fit, newdata=list(x1=0, x2=0, sex=1))
# -0.1497816

## by predict function
predict(fit, newdata=list(x1=1, x2=2, sex=0))
# 0.009588167
predict(fit, newdata=list(x1=1, x2=2, sex=1))
# 0.6063845

## by hand
lambda_sex1 + coef(fit)[1]*1 + coef(fit)[2]*2
# 0.009588167
lambda_sex2 + coef(fit)[1]*1 + coef(fit)[2]*2
# 0.6063845
</code></pre>
<p>Finally, to calculate &quot;a&quot; linear predictor for an unknown <code>sex</code>:</p>
<pre><code># prevented by predict function
predict(fit, newdata=list(x1=1, x2=2, sex=2))
#Error in model.frame.default(data = list(x1 = 1, x2 = 2, sex = 2), formula = ~x1 +  : 
# factor strata(sex) has new level sex=2

# by hand, assuming lambda for sex 2 is 0
0 + coef(fit)[1]*1 + coef(fit)[2]*2
# 0.7561661
</code></pre>

CC BY-SA 4.0
2020-07-23T13:12:40.710Z
2020-07-23T13:38:52.647Z
2020-07-23T13:38:52.647Z
Ryan SY Kwan
<p>I have fitted a cox model on a pooled dataset of multiple studies, say studies A, B, and C. As these studies have different baseline hazard functions, I stratified for 'study' in the model.</p>
<p>Like this:</p>
<pre><code>df$SurvObj &lt;- with(df, Surv(event_rd, event == 1))

fit &lt;- coxph(SurvObj ~ cov1 + cov2 + cov3 + cov4 + strata(study),data=df)
</code></pre>
<p>Now, I want to assess whether I can use the beta coefficients from the above model to predict the event probability in a new study, study D.
The baseline hazard in study D is different from those in studies A, B, and C.</p>
<p>When I have</p>
<pre><code>predict(fit,type=&quot;lp&quot;,newdata=studyD,reference=&quot;strata&quot;)
</code></pre>
<p>I get this error:</p>
<blockquote>
<p>Error in model.frame.default(data = studyD, formula = ~hba1c + sbp +
: factor strata(study) has new levels study=4</p>
</blockquote>
<p>Why does R require that I match the strata of study D with those in study A, B, and/or C when I only want to output the linear predictor (<code>type=&quot;lp&quot;</code>)? As the linear predictor is independent of the baseline hazard function in this case.</p>
<p>I suppose it should be possible to extract the linear predictor for individuals in study D and then manually calculate the event probability using the baseline hazard of study D.</p>
<p>Does anybody know how to do this?<br />
I could not find any question like this so maybe I am thinking the wrong way.</p>

CC BY-SA 4.0
2019-02-04T12:21:55.597Z
I would recommend trying out Stats.SE (aka CrossValidated) as some folks there might be of more help.
posdef
<r>
Output the linear predictor from a stratified cox model?
# Output the linear predictor from a stratified cox model?

(Asked by: username_0 on 2019-01-26)

I have fitted a cox model on a pooled dataset of multiple studies, say studies A, B, and C. As these studies have different baseline hazard functions, I stratified for 'study' in the model.
Like this:
df$SurvObj <- with(df, Surv(event_rd, event == 1))

fit <- coxph(SurvObj ~ cov1 + cov2 + cov3 + cov4 + strata(study),data=df)

Now, I want to assess whether I can use the beta coefficients from the above model to predict the event probability in a new study, study D.
The baseline hazard in study D is different from those in studies A, B, and C.
When I have
predict(fit,type="lp",newdata=studyD,reference="strata")

I get this error:

Error in model.frame.default(data = studyD, formula = ~hba1c + sbp +
: factor strata(study) has new levels study=4

Why does R require that I match the strata of study D with those in study A, B, and/or C when I only want to output the linear predictor (type="lp")? As the linear predictor is independent of the baseline hazard function in this case.
I suppose it should be possible to extract the linear predictor for individuals in study D and then manually calculate the event probability using the baseline hazard of study D.
Does anybody know how to do this?
I could not find any question like this so maybe I am thinking the wrong way.

**Tags:** r
> **Comments:**
> **username_1**: I would recommend trying out Stats.SE (aka CrossValidated) as some folks there might be of more help.  

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## 1 answers

### Answer by username_2 on 2020-07-23. Score: 3

Unlike wikipedia, the linear predictor (LP) in coxph is made dependent from the baseline hazard function $\lambda$. For example, if a model has two predictors x1, x2 stratified by sex, the linear predictor would be like:
$$LP = log (  \lambda _{sex}) + \beta _{1}x _{1}+\beta _{2}x _{2}$$
Where the $log(\lambda _{sex})$ can be obtained by predict() with x1=x2=0.
We could calculate the LP without strata information (sex) by hand using the $\beta _{1} \, \beta _{2}$ provided by coef(fit) and a customised $\lambda _{unkown \, sex}$.
Below is a simulation comparing the results from predict() and by hand coef(fit).
## Toy data
df <- list(time=c(4,3,1,1,2,2,3), 
          status=c(1,1,1,0,1,1,0), 
          x1=c(0,2,1,1,1,0,0), 
          x2=1:7,
          sex=c(0,0,0,0,1,1,1)) 

## Fit a stratified model 
fit <- coxph(Surv(time, status) ~ x1 + x2 + strata(sex), df) 
coef(fit)
#         x1          x2 
# 0.79451881 -0.01917633

## Baseline linear predictor
lambda_sex1 <- predict(fit, newdata=list(x1=0, x2=0, sex=0))
# -0.746578
lambda_sex2 <- predict(fit, newdata=list(x1=0, x2=0, sex=1))
# -0.1497816

## by predict function
predict(fit, newdata=list(x1=1, x2=2, sex=0))
# 0.009588167
predict(fit, newdata=list(x1=1, x2=2, sex=1))
# 0.6063845

## by hand
lambda_sex1 + coef(fit)[1]*1 + coef(fit)[2]*2
# 0.009588167
lambda_sex2 + coef(fit)[1]*1 + coef(fit)[2]*2
# 0.6063845

Finally, to calculate "a" linear predictor for an unknown sex:
# prevented by predict function
predict(fit, newdata=list(x1=1, x2=2, sex=2))
#Error in model.frame.default(data = list(x1 = 1, x2 = 2, sex = 2), formula = ~x1 +  : 
# factor strata(sex) has new level sex=2

# by hand, assuming lambda for sex 2 is 0
0 + coef(fit)[1]*1 + coef(fit)[2]*2
# 0.7561661


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