How to split a data frame based on one variable and create cross table separetely between other two variable...











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My dataset: df 



PID<-c(1,2,3,4,5,6,7,8,9)  

gender<-c(1,1,0,1,0,0,0,1,1)  

smoking<-c(1,1,0,0,0,0,1,0,1)  

disease<-c(1,0,0,1,1,1,0,1,0)  

BMI<-c(24,23,21,28,29,21,18,19,16)    

 df<-data.frame(PID, gender, smoking, disease, BMI)


I want to split this dataset based on gender. Then develop crosstab between smoking and disease. How to do this?
Expected outcome(first question):

Gender:1

crosstab between smoking and disease



Gender:2

Crosstab between smoking and disease.



Expected outcome (second question):



Gender:1

mean of BMI



Gender:2

mean of BMI






r







share|improve this question
























  • Have you tried anything yet? I'd think xtabs and aggregate should answer both questions.
    – r2evans
    Nov 5 at 2:28















up vote
-1
down vote

favorite












My dataset: df 



PID<-c(1,2,3,4,5,6,7,8,9)  

gender<-c(1,1,0,1,0,0,0,1,1)  

smoking<-c(1,1,0,0,0,0,1,0,1)  

disease<-c(1,0,0,1,1,1,0,1,0)  

BMI<-c(24,23,21,28,29,21,18,19,16)    

 df<-data.frame(PID, gender, smoking, disease, BMI)


I want to split this dataset based on gender. Then develop crosstab between smoking and disease. How to do this?
Expected outcome(first question):

Gender:1

crosstab between smoking and disease



Gender:2

Crosstab between smoking and disease.



Expected outcome (second question):



Gender:1

mean of BMI



Gender:2

mean of BMI






r







share|improve this question
























  • Have you tried anything yet? I'd think xtabs and aggregate should answer both questions.
    – r2evans
    Nov 5 at 2:28













up vote
-1
down vote

favorite









up vote
-1
down vote

favorite











My dataset: df 



PID<-c(1,2,3,4,5,6,7,8,9)  

gender<-c(1,1,0,1,0,0,0,1,1)  

smoking<-c(1,1,0,0,0,0,1,0,1)  

disease<-c(1,0,0,1,1,1,0,1,0)  

BMI<-c(24,23,21,28,29,21,18,19,16)    

 df<-data.frame(PID, gender, smoking, disease, BMI)


I want to split this dataset based on gender. Then develop crosstab between smoking and disease. How to do this?
Expected outcome(first question):

Gender:1

crosstab between smoking and disease



Gender:2

Crosstab between smoking and disease.



Expected outcome (second question):



Gender:1

mean of BMI



Gender:2

mean of BMI






r







share|improve this question















My dataset: df 



PID<-c(1,2,3,4,5,6,7,8,9)  

gender<-c(1,1,0,1,0,0,0,1,1)  

smoking<-c(1,1,0,0,0,0,1,0,1)  

disease<-c(1,0,0,1,1,1,0,1,0)  

BMI<-c(24,23,21,28,29,21,18,19,16)    

 df<-data.frame(PID, gender, smoking, disease, BMI)


I want to split this dataset based on gender. Then develop crosstab between smoking and disease. How to do this?
Expected outcome(first question):

Gender:1

crosstab between smoking and disease



Gender:2

Crosstab between smoking and disease.



Expected outcome (second question):



Gender:1

mean of BMI



Gender:2

mean of BMI






r




r






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share|improve this question













share|improve this question




share|improve this question








edited Nov 5 at 2:31

























asked Nov 5 at 2:15









Bikram Adhitya Adhikari

154




154












  • Have you tried anything yet? I'd think xtabs and aggregate should answer both questions.
    – r2evans
    Nov 5 at 2:28


















  • Have you tried anything yet? I'd think xtabs and aggregate should answer both questions.
    – r2evans
    Nov 5 at 2:28
















Have you tried anything yet? I'd think xtabs and aggregate should answer both questions.
– r2evans
Nov 5 at 2:28




Have you tried anything yet? I'd think xtabs and aggregate should answer both questions.
– r2evans
Nov 5 at 2:28












2 Answers
2






active

oldest

votes

















up vote
2
down vote













No need for external packages:



xtabs(~smoking+disease+gender,data=df)

# , , gender = 0

#        disease

# smoking 0 1

#       0 1 2

#       1 1 0

# , , gender = 1

#        disease

# smoking 0 1

#       0 0 2

#       1 2 1

aggregate(df$BMI, list(gender=df$gender), FUN=mean)

#   gender     x

# 1      0 22.25

# 2      1 22.00


Similarly (thanks thelatemail):



aggregate(BMI ~ gender, data = df, FUN=mean)





share|improve this answer



















  • 1




    aggregate(BMI ~ gender, data=df, FUN=mean) might be a bit more readable and keep the consistent formula interface. Or even aggregate(df["BMI"], df["gender"], FUN=mean) if the formulas are not to your taste.
    – thelatemail
    Nov 5 at 3:56




















up vote
1
down vote













Here is a possible way for the first question using dplyr:



library(dplyr)

library(magrittr)





> df %>% split(gender) %>% lapply(function(x) tab=xtabs(gender~smoking+disease, data=x))

$`0`

       disease

smoking 0 1

      0 0 0

      1 0 0



$`1`

       disease

smoking 0 1

      0 0 2

      1 2 1





share|improve this answer





















  • You really don't need dlpyr here ... xtabs(~smoking+disease+gender,data=df) does effectively the same thing
    – r2evans
    Nov 5 at 3:19











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote













No need for external packages:



xtabs(~smoking+disease+gender,data=df)

# , , gender = 0

#        disease

# smoking 0 1

#       0 1 2

#       1 1 0

# , , gender = 1

#        disease

# smoking 0 1

#       0 0 2

#       1 2 1

aggregate(df$BMI, list(gender=df$gender), FUN=mean)

#   gender     x

# 1      0 22.25

# 2      1 22.00


Similarly (thanks thelatemail):



aggregate(BMI ~ gender, data = df, FUN=mean)





share|improve this answer



















  • 1




    aggregate(BMI ~ gender, data=df, FUN=mean) might be a bit more readable and keep the consistent formula interface. Or even aggregate(df["BMI"], df["gender"], FUN=mean) if the formulas are not to your taste.
    – thelatemail
    Nov 5 at 3:56

















up vote
2
down vote













No need for external packages:



xtabs(~smoking+disease+gender,data=df)

# , , gender = 0

#        disease

# smoking 0 1

#       0 1 2

#       1 1 0

# , , gender = 1

#        disease

# smoking 0 1

#       0 0 2

#       1 2 1

aggregate(df$BMI, list(gender=df$gender), FUN=mean)

#   gender     x

# 1      0 22.25

# 2      1 22.00


Similarly (thanks thelatemail):



aggregate(BMI ~ gender, data = df, FUN=mean)





share|improve this answer



















  • 1




    aggregate(BMI ~ gender, data=df, FUN=mean) might be a bit more readable and keep the consistent formula interface. Or even aggregate(df["BMI"], df["gender"], FUN=mean) if the formulas are not to your taste.
    – thelatemail
    Nov 5 at 3:56















up vote
2
down vote










up vote
2
down vote









No need for external packages:



xtabs(~smoking+disease+gender,data=df)

# , , gender = 0

#        disease

# smoking 0 1

#       0 1 2

#       1 1 0

# , , gender = 1

#        disease

# smoking 0 1

#       0 0 2

#       1 2 1

aggregate(df$BMI, list(gender=df$gender), FUN=mean)

#   gender     x

# 1      0 22.25

# 2      1 22.00


Similarly (thanks thelatemail):



aggregate(BMI ~ gender, data = df, FUN=mean)





share|improve this answer














No need for external packages:



xtabs(~smoking+disease+gender,data=df)

# , , gender = 0

#        disease

# smoking 0 1

#       0 1 2

#       1 1 0

# , , gender = 1

#        disease

# smoking 0 1

#       0 0 2

#       1 2 1

aggregate(df$BMI, list(gender=df$gender), FUN=mean)

#   gender     x

# 1      0 22.25

# 2      1 22.00


Similarly (thanks thelatemail):



aggregate(BMI ~ gender, data = df, FUN=mean)






share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 5 at 4:03

























answered Nov 5 at 3:19









r2evans

23.7k32856




23.7k32856








  • 1




    aggregate(BMI ~ gender, data=df, FUN=mean) might be a bit more readable and keep the consistent formula interface. Or even aggregate(df["BMI"], df["gender"], FUN=mean) if the formulas are not to your taste.
    – thelatemail
    Nov 5 at 3:56
















  • 1




    aggregate(BMI ~ gender, data=df, FUN=mean) might be a bit more readable and keep the consistent formula interface. Or even aggregate(df["BMI"], df["gender"], FUN=mean) if the formulas are not to your taste.
    – thelatemail
    Nov 5 at 3:56










1




1




aggregate(BMI ~ gender, data=df, FUN=mean) might be a bit more readable and keep the consistent formula interface. Or even aggregate(df["BMI"], df["gender"], FUN=mean) if the formulas are not to your taste.
– thelatemail
Nov 5 at 3:56






aggregate(BMI ~ gender, data=df, FUN=mean) might be a bit more readable and keep the consistent formula interface. Or even aggregate(df["BMI"], df["gender"], FUN=mean) if the formulas are not to your taste.
– thelatemail
Nov 5 at 3:56














up vote
1
down vote













Here is a possible way for the first question using dplyr:



library(dplyr)

library(magrittr)





> df %>% split(gender) %>% lapply(function(x) tab=xtabs(gender~smoking+disease, data=x))

$`0`

       disease

smoking 0 1

      0 0 0

      1 0 0



$`1`

       disease

smoking 0 1

      0 0 2

      1 2 1





share|improve this answer





















  • You really don't need dlpyr here ... xtabs(~smoking+disease+gender,data=df) does effectively the same thing
    – r2evans
    Nov 5 at 3:19















up vote
1
down vote













Here is a possible way for the first question using dplyr:



library(dplyr)

library(magrittr)





> df %>% split(gender) %>% lapply(function(x) tab=xtabs(gender~smoking+disease, data=x))

$`0`

       disease

smoking 0 1

      0 0 0

      1 0 0



$`1`

       disease

smoking 0 1

      0 0 2

      1 2 1





share|improve this answer





















  • You really don't need dlpyr here ... xtabs(~smoking+disease+gender,data=df) does effectively the same thing
    – r2evans
    Nov 5 at 3:19













up vote
1
down vote










up vote
1
down vote









Here is a possible way for the first question using dplyr:



library(dplyr)

library(magrittr)





> df %>% split(gender) %>% lapply(function(x) tab=xtabs(gender~smoking+disease, data=x))

$`0`

       disease

smoking 0 1

      0 0 0

      1 0 0



$`1`

       disease

smoking 0 1

      0 0 2

      1 2 1





share|improve this answer












Here is a possible way for the first question using dplyr:



library(dplyr)

library(magrittr)





> df %>% split(gender) %>% lapply(function(x) tab=xtabs(gender~smoking+disease, data=x))

$`0`

       disease

smoking 0 1

      0 0 0

      1 0 0



$`1`

       disease

smoking 0 1

      0 0 2

      1 2 1






share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 5 at 3:06









mysteRious

1,9702512




1,9702512












  • You really don't need dlpyr here ... xtabs(~smoking+disease+gender,data=df) does effectively the same thing
    – r2evans
    Nov 5 at 3:19


















  • You really don't need dlpyr here ... xtabs(~smoking+disease+gender,data=df) does effectively the same thing
    – r2evans
    Nov 5 at 3:19
















You really don't need dlpyr here ... xtabs(~smoking+disease+gender,data=df) does effectively the same thing
– r2evans
Nov 5 at 3:19




You really don't need dlpyr here ... xtabs(~smoking+disease+gender,data=df) does effectively the same thing
– r2evans
Nov 5 at 3:19


















 

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