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Given a hazard ratio($HR$) as $$𝐻𝑅_{b}=β„Žπ‘Žπ‘§π‘Žπ‘Ÿπ‘‘(b)/β„Žπ‘Žπ‘§π‘Žπ‘Ÿπ‘‘(a)$$ and another hazard ratio as $$𝐻𝑅_{c}=β„Žπ‘Žπ‘§π‘Žπ‘Ÿπ‘‘(c)/β„Žπ‘Žπ‘§π‘Žπ‘Ÿπ‘‘(a)$$ How can we obtain the hazard ratio between $b$ and $c$ from $HR_{b}$ and $HR_{c}$ ?

Here $hazard(a),hazard(b),hazard(c)$ are hazards for three subgroups consisting of samples of a bigger data-set on which cluster analysis was applied to find the 3 subgroups(of samples). The model was decribed here before

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    $\begingroup$ Write out the full models (you're probably conditioning on something, e.g. covariates) and sample characteristics/study design, e.g., are HRb and HRc estimated using the same dataset so that a single protocol is in place? $\endgroup$ Apr 6, 2019 at 12:26
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    $\begingroup$ If my answer(s) clarify your questions consider marking them as answered. $\endgroup$
    – user213325
    Apr 6, 2019 at 12:37
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    $\begingroup$ Using cluster analysis to find groups to test is almost always problematic. You either need to use the continuous distance from cluster centers for each observations, or directly analyze the measurements that went into the clustering and drop the clustering. $\endgroup$ Apr 9, 2019 at 4:13
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    $\begingroup$ I wasn't clear on whether you were clustering observations or clustering variables. It seems you are clustering observations, which is the more problematic situation. Clustering, at least the kind I'm familiar with, groups observations but within each cluster there is too much heterogeneity of the observations to consider them as a homogeneous group as you are doing. Continuous distances or non-clustering methods will fix this. (Imagine 2 clusters as touching circles; a subject at the edge of one may be more like obs. in the other cluster than in her own cluster.) $\endgroup$ Apr 9, 2019 at 12:16
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    $\begingroup$ I don't have a reference at my fingertips for the distance from cluster center, but it is motivated by the need to take into account heterogeneity within clusters, i.e., clusters don't have a radius of zero. For non-cluster methods see my RMS book and course notes. $\endgroup$ Apr 9, 2019 at 19:03

1 Answer 1

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You can do it with HR(c)/HR(b)

The great thing about R is that you can try it if you are not sure. Like this here:

# firstly, load some data
data(lung)

# creating some factor
# number of each cluster (I use 3 clusters)
nk <- nrow(lung)/3
# create some factor
lung$cluster <- factor(c(rep("cluster1", nk), rep("cluster2", 
nk), rep("cluster3", nk)))

# fit models
# Here reference level is cluster1
m1 <- coxph(Surv(time, status) ~ cluster, data=lung)

# with cluster 2 as reference
# change reference level to cluster2
lung$cluster <- relevel(lung$cluster, "cluster2")
# fit a model with cluster2 as reference
m2 <- coxph(Surv(time, status) ~ cluster, data=lung)

# calculating hazard ratio for cluater2 vs. cluster3 from m1
exp(m1$coefficients["clustercluster3"]) / 
  exp(m1$coefficients["clustercluster2"])
0.8416011

# comparing it to the hazard ratio cluster2 vs. cluster3. from model fit
exp(m2$coefficients["clustercluster3"])
0.8416011
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