I am trying to get an Hazard Ratio for 4 survival curves according quartiles of delta_mon1_baselone_to_3d using this code:
quantile <- df$delta_mon1_baseline_to_d3
cox <- coxph(Surv(mace_months_date_vs_date_sample, mace) ~ findInterval(quantile, quantile(quantile, na.rm = TRUE)[-5]), data = df)
summary(cox)
But I only get one hazard ratio.
I have seen that I have to convert the predictor column delta_mon1_baselone_to_3d as factor using this code:
df$quantile <- as.factor(df$quantile)
but cox regression doesn't work after converting.
Converting directly df$quartile_mon1_delta3d <- as.factor(df$quartile_mon1_delta3d) gives me the same feedback.
Here are my data:
ID age sex mace mace_months_date_vs_date_sample trop egfr dm smoke delta_mon1_baseline_to_d3
1 44 52 1 1 30 2600 56 1 0 -822.
2 32 66 1 0 73 1710 90 1 0 -562.
3 20 56 1 1 5 NA 75 0 1 -502.
4 17 44 1 0 77 840 71 0 0 -389.
5 52 49 1 0 74 1740 71 0 1 -372.
6 57 58 1 0 74 5010 68 0 1 -308.
7 79 68 1 0 45 776 90 0 0 -284.
8 14 74 1 1 6 7120 78 0 0 -279.
9 223 63 1 0 46 4281 90 0 0 -218.
10 56 43 1 0 70 1360 90 1 0 -173.
11 50 54 1 0 70 15300 90 0 1 -163.
12 31 47 0 0 72 6490 77 1 1 -95.7
13 35 47 1 0 77 NA 83 0 0 -71.0
14 36 64 1 1 5 15940 52 0 1 -69.7
15 15 65 1 1 43 6300 49 1 0 -69.6
16 12 57 1 0 71 6020 88 0 1 -66.5
17 43 59 0 0 74 2100 84 0 1 -58.8
18 22 46 1 0 77 5330 88 0 1 -29.3
19 54 59 1 0 71 1500 81 1 1 -25.7
20 26 66 1 0 77 500 51 0 0 -12.5
21 29 73 0 0 77 NA 51 0 0 -2.99
22 25 54 1 0 73 1080 87 0 0 2.81
23 39 54 1 0 74 990 77 0 0 32.9
24 47 62 1 0 69 1420 85 0 0 33.0
25 49 54 1 1 28 NA 76 1 0 44.1
26 24 47 1 0 77 2390 90 0 1 47.7
27 45 51 0 0 73 3710 65 0 1 55.9
28 30 73 0 0 68 3340 48 1 1 117.
29 16 57 1 0 73 180 99 0 1 131.
30 55 47 1 0 70 NA 90 0 1 131.
31 37 81 1 0 74 NA 99 1 1 147.
32 21 46 1 0 75 3600 87 0 1 153.
33 60 72 1 0 76 470 62 0 0 160.
34 18 56 1 0 69 6390 90 0 1 165.
35 13 53 1 0 69 1970 87 1 1 180.
36 19 66 1 0 78 9320 59 0 0 180.
37 33 59 1 0 69 2260 79 0 1 193.
38 139 39 0 0 58 NA 90 0 1 209.
39 38 55 1 0 78 3930 90 1 0 244.
40 27 28 1 0 71 6440 90 0 1 248.
41 58 36 1 0 76 NA 78 1 1 327.
42 61 48 1 0 76 4470 90 0 1 336.
43 42 38 1 0 69 1800 69 0 1 375.
44 28 76 1 0 71 40 90 1 1 419.
Any help would be great! Thanks
CodePudding user response：
If I understand you correctly, you want to use the quantile as a categorical dependent variable in the Cox PH model. I can't recreate your error, but this produces coefficient estimates for each quantile using the first quantile as the reference:
# Determine quantiles
df$quantile <- cut(df$delta_mon1_baseline_to_d3, quantile(df$delta_mon1_baseline_to_d3))
# Run model (per OP model)
cox <- survival::coxph(survival::Surv(mace_months_date_vs_date_sample, mace) ~ df$quantile, data = df)
# survival::coxph(formula = survival::Surv(mace_months_date_vs_date_sample,
# mace) ~ df$quantile, data = df)
#
# coef exp(coef) se(coef) z p
# df$quantile(-113,17.9] -1.490e-01 8.616e-01 1.000e 00 -0.149 0.882
# df$quantile(17.9,161] -8.902e-01 4.106e-01 1.225e 00 -0.727 0.467
# df$quantile(161,419] -1.990e 01 2.274e-09 1.331e 04 -0.001 0.999
CodePudding user response：
From the data you have provided, you can do:
library(survival)
df$quantile <- df$delta_mon1_baseline_to_d3
df$stratum <- findInterval(df$quantile, quantile(df$quantile, na.rm = TRUE)[-5])
df$stratum <- factor(df$stratum)
cox <- coxph(Surv(mace_months_date_vs_date_sample, mace) ~ stratum, data = df)
summary(cox)
#> Call:
#> coxph(formula = Surv(mace_months_date_vs_date_sample, mace) ~
#> stratum, data = df)
#>
#> n= 44, number of events= 6
#>
#> coef exp(coef) se(coef) z Pr(>|z|)
#> stratum2 -4.780e-01 6.201e-01 9.135e-01 -0.523 0.601
#> stratum3 -1.211e 00 2.980e-01 1.155e 00 -1.048 0.295
#> stratum4 -2.015e 01 1.767e-09 1.277e 04 -0.002 0.999
#>
#> exp(coef) exp(-coef) lower .95 upper .95
#> stratum2 6.201e-01 1.613e 00 0.10348 3.715
#> stratum3 2.980e-01 3.356e 00 0.03097 2.867
#> stratum4 1.767e-09 5.661e 08 0.00000 Inf
#>
#> Concordance= 0.731 (se = 0.08 )
#> Likelihood ratio test= 4.99 on 3 df, p=0.2
#> Wald test = 1.14 on 3 df, p=0.8
#> Score (logrank) test = 3.86 on 3 df, p=0.3
This shows 3 hazard ratios (comparisons of quantiles 2-4 to quantile 1).