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Hi Experts
Need to add the following as calcaluated column measure into my pbix file (see images below) i am trying to populate columns I, J and k into my pbix file. The formulas used in excel are shown below.
Sample file.
link: https://www.dropbox.com/s/47fyhkj5euzw02q/Kaplan_Miers_v1.2.pbix?dl=0
Solved! Go to Solution.
Hey @Anonymous , I received a request regarding confidence intervals with Kaplan Meier survival curves from @Anonymous and so I was working on it and came up with the following answer:
OK, following along from this:
And if you are following my article here:
Then in your KMDeptDaysAll table, create this measure:
ci_alpha = 0.05
This is your confidence interval, in this case 95%
And then create these columns:
ci_S(t) = VAR __table = FILTER(ALL(KMDeptDaysAll),[Department] = EARLIER([Department]) && [Days] <= EARLIER([Days])) RETURN PRODUCTX(__table,[1-e_i/d_i]) ci_se_1 = VAR __calc = [e_i] / ([d_i] * ([d_i] - [e_i])) RETURN IF(ISERROR(SQRT(__calc)),BLANK(),[ci_S(t)] * SQRT(__calc)) ci_se = VAR __table = FILTER(ALL(KMDeptDaysAll),[Department] = EARLIER([Department]) && [Days] <= EARLIER([Days])) VAR __sum = SUMX(__table,[ci_se_1]) RETURN [ci_S(t)] * __sum ci_lower = [ci_S(t)] ^ IF(ISERROR(EXP(NORM.S.INV([ci_alpha]/2)/IF(ISERROR(LN([ci_S(t)])),BLANK(),LN([ci_S(t)])*[ci_se]/[ci_S(t)]))),BLANK(),EXP(NORM.S.INV([ci_alpha]/2)/IF(ISERROR(LN([ci_S(t)])),BLANK(),LN([ci_S(t)])*[ci_se]/[ci_S(t)]))) ci_upper = [ci_S(t)] ^ IF(ISERROR(EXP(-NORM.S.INV([ci_alpha]/2)/IF(ISERROR(LN([ci_S(t)])),BLANK(),LN([ci_S(t)])*[ci_se]/[ci_S(t)]))),BLANK(),EXP(-NORM.S.INV([ci_alpha]/2)/IF(ISERROR(LN([ci_S(t)])),BLANK(),LN([ci_S(t)])*[ci_se]/[ci_S(t)])))
Can't be certain this is all 100% correct as the original article I was working from didn't have confidence intervals. Will try to do some verification work on this.
Hey @Anonymous , I received a request regarding confidence intervals with Kaplan Meier survival curves from @Anonymous and so I was working on it and came up with the following answer:
OK, following along from this:
And if you are following my article here:
Then in your KMDeptDaysAll table, create this measure:
ci_alpha = 0.05
This is your confidence interval, in this case 95%
And then create these columns:
ci_S(t) = VAR __table = FILTER(ALL(KMDeptDaysAll),[Department] = EARLIER([Department]) && [Days] <= EARLIER([Days])) RETURN PRODUCTX(__table,[1-e_i/d_i]) ci_se_1 = VAR __calc = [e_i] / ([d_i] * ([d_i] - [e_i])) RETURN IF(ISERROR(SQRT(__calc)),BLANK(),[ci_S(t)] * SQRT(__calc)) ci_se = VAR __table = FILTER(ALL(KMDeptDaysAll),[Department] = EARLIER([Department]) && [Days] <= EARLIER([Days])) VAR __sum = SUMX(__table,[ci_se_1]) RETURN [ci_S(t)] * __sum ci_lower = [ci_S(t)] ^ IF(ISERROR(EXP(NORM.S.INV([ci_alpha]/2)/IF(ISERROR(LN([ci_S(t)])),BLANK(),LN([ci_S(t)])*[ci_se]/[ci_S(t)]))),BLANK(),EXP(NORM.S.INV([ci_alpha]/2)/IF(ISERROR(LN([ci_S(t)])),BLANK(),LN([ci_S(t)])*[ci_se]/[ci_S(t)]))) ci_upper = [ci_S(t)] ^ IF(ISERROR(EXP(-NORM.S.INV([ci_alpha]/2)/IF(ISERROR(LN([ci_S(t)])),BLANK(),LN([ci_S(t)])*[ci_se]/[ci_S(t)]))),BLANK(),EXP(-NORM.S.INV([ci_alpha]/2)/IF(ISERROR(LN([ci_S(t)])),BLANK(),LN([ci_S(t)])*[ci_se]/[ci_S(t)])))
Can't be certain this is all 100% correct as the original article I was working from didn't have confidence intervals. Will try to do some verification work on this.
Hi @Anonymous ,
In your sample, I found there are some values like negative number and ∞. They cause that the SQRT function returns an error. You could use the function of IF to handle them firstly.
Best Regards,
Xue Ding
If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.
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