# How to run Logistic Regression on Aggregate Data in R

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We will provide an example of how you can run a logistic regression in R when the data are grouped. Let’s provide some random sample data of 200 observations.

library(tidyverse) set.seed(5) df<-tibble(Gender = as.factor(sample(c("m","f"), 200, replace = TRUE, prob=c(0.6,0.4))), Age_Group = as.factor(sample(c("[<30]","[30-65]", "[65+]"), 200, replace = TRUE, prob=c(0.3,0.6,0.1))), Response = rbinom(200, 1, prob = 0.2)) df

**Output:**

# A tibble: 200 x 3 Gender Age_Group Response1 f [65+] 0 2 m [30-65] 0 3 m [65+] 0 4 m [30-65] 0 5 m [<30] 0 6 m [<30] 0 7 m [30-65] 0 8 m [30-65] 0 9 f [<30] 1 10 f [<30] 0 # ... with 190 more rows

## Logistic Regression on Non-Aggregate Data

The logistic regression model is the following:

model1<-glm(Response ~ Gender+Age_Group, data = df, family = binomial("logit")) summary(model1)

**Output:**

Call: glm(formula = Response ~ Gender + Age_Group, family = binomial("logit"), data = df) Deviance Residuals: Min 1Q Median 3Q Max -0.7039 -0.6246 -0.6094 -0.5677 1.9754 Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -1.32296 0.40899 -3.235 0.00122 ** Genderm 0.05402 0.38041 0.142 0.88707 Age_Group[30-65] -0.26642 0.42010 -0.634 0.52596 Age_Group[65+] -0.47482 0.59460 -0.799 0.42455 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 (Dispersion parameter for binomial family taken to be 1) Null deviance: 188.56 on 199 degrees of freedom Residual deviance: 187.83 on 196 degrees of freedom AIC: 195.83 Number of Fisher Scoring iterations: 4

## Logistic Regression on Aggregate Data

Assume now that you have received the data in an aggregated form and you were asked to run logistic regression. First, we need to generate the aggregate data.

df_agg<-df%>%group_by(Gender, Age_Group)%>%summarise(Impressions=n(), Responses=sum(Response))%>% ungroup()%>%mutate(RR=Responses/Impressions) df_agg

**Output:**

# A tibble: 6 x 5 Gender Age_Group Impressions Responses RR1 f [<30] 21 6 0.286 2 f [30-65] 49 7 0.143 3 f [65+] 9 1 0.111 4 m [<30] 30 5 0.167 5 m [30-65] 66 13 0.197 6 m [65+] 25 4 0.16

Below we will represent three different solutions.

**Logistic Regression with Weights**

m2<-glm(RR ~ Gender+Age_Group, data=df_agg, weights = Impressions, family = binomial("logit")) summary(m2)

**Output:**

Call: glm(formula = RR ~ Gender + Age_Group, family = binomial("logit"), data = df_agg, weights = Impressions) Deviance Residuals: 1 2 3 4 5 6 0.8160 -0.5077 -0.2754 -0.7213 0.4145 0.1553 Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -1.32296 0.40899 -3.235 0.00122 ** Genderm 0.05402 0.38042 0.142 0.88707 Age_Group[30-65] -0.26642 0.42010 -0.634 0.52596 Age_Group[65+] -0.47482 0.59460 -0.799 0.42455 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 (Dispersion parameter for binomial family taken to be 1) Null deviance: 2.4477 on 5 degrees of freedom Residual deviance: 1.7157 on 2 degrees of freedom AIC: 29.167 Number of Fisher Scoring iterations: 4

**Logistic Regression with cbind**

We will need to create another column called of the `No Responses`

and then we can use the `cbind`

:

df_agg$No_Responses <- df_agg$Impressions- df_agg$Responses m3<-glm(cbind(Responses, No_Responses) ~ Gender+Age_Group, data=df_agg, family = binomial("logit")) summary(m3)

**Output:**

Call: glm(formula = cbind(Responses, No_Responses) ~ Gender + Age_Group, family = binomial("logit"), data = df_agg) Deviance Residuals: 1 2 3 4 5 6 0.8160 -0.5077 -0.2754 -0.7213 0.4145 0.1553 Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -1.32296 0.40899 -3.235 0.00122 ** Genderm 0.05402 0.38042 0.142 0.88707 Age_Group[30-65] -0.26642 0.42010 -0.634 0.52596 Age_Group[65+] -0.47482 0.59460 -0.799 0.42455 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 (Dispersion parameter for binomial family taken to be 1) Null deviance: 2.4477 on 5 degrees of freedom Residual deviance: 1.7157 on 2 degrees of freedom AIC: 29.167 Number of Fisher Scoring iterations: 4

**Expand the Aggregate Data**

Finally, another approach will be to transform the aggregate data to the binary form of 0 and 1. Let’s do it:

df2 <- df_agg %>% mutate(New_Response = map2(Responses, Impressions, ~ c(rep(1, .x), rep(0, .y - .x))))%>%unnest(cols = c(New_Response)) df2

**Output:**

# A tibble: 200 x 7 Gender Age_Group Impressions Responses RR No_Responses New_Response1 f [<30] 21 6 0.286 15 1 2 f [<30] 21 6 0.286 15 1 3 f [<30] 21 6 0.286 15 1 4 f [<30] 21 6 0.286 15 1 5 f [<30] 21 6 0.286 15 1 6 f [<30] 21 6 0.286 15 1 7 f [<30] 21 6 0.286 15 0 8 f [<30] 21 6 0.286 15 0 9 f [<30] 21 6 0.286 15 0 10 f [<30] 21 6 0.286 15 0 # ... with 190 more rows

And now we can run similarly with what we did at the beginning.

model4<-glm(New_Response ~ Gender+Age_Group, data = df2, family = binomial("logit")) summary(model4)

**Output:**

Call: glm(formula = New_Response ~ Gender + Age_Group, family = binomial("logit"), data = df2) Deviance Residuals: Min 1Q Median 3Q Max -0.7039 -0.6246 -0.6094 -0.5677 1.9754 Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -1.32296 0.40899 -3.235 0.00122 ** Genderm 0.05402 0.38041 0.142 0.88707 Age_Group[30-65] -0.26642 0.42010 -0.634 0.52596 Age_Group[65+] -0.47482 0.59460 -0.799 0.42455 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 (Dispersion parameter for binomial family taken to be 1) Null deviance: 188.56 on 199 degrees of freedom Residual deviance: 187.83 on 196 degrees of freedom AIC: 195.83 Number of Fisher Scoring iterations: 4

## The Takeaway

With all 4 models, we came up with the same coefficients and p-values. However, in the aggregate form, we get different output regarding the deviance and the AIC score compared to the binary form.

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