How to loop a function over columns in many dataframes in r

user_11729654 0
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I have a number of dataframes (96) that have columns of 0s and 1s. If there is more than one "1" in any one column of the dataframes, I want to replace the 1s with a equal fractions so that the sum of the values in the column is 1, as shown in the code below. 

v1 <- c(0, 1, 0, 1, 1, 0)
v2 <- c(0, 0, 1, 0, 0, 0)
v3 <- c(0, 0, 1, 1, 0, 0)
df1 <- data.frame(v1, v2, v3)
df2 <-data.frame(v3, v3, v1)
df3 <- data.frame(v1, v3, v1)

new.df1 <- t(apply(df1, 2, FUN = function(x) {
  if(sum(x==1, na.rm=TRUE) ==2)  replace(x, x==1, 0.5)
  else if (sum(x==1, na.rm=TRUE)==3) replace( x, x==1, 1/3) 
  else x}))

new.df2 <- t(apply(df2, 2, FUN = function(x) {
  if(sum(x==1, na.rm=TRUE) ==2)  replace(x, x==1, 0.5)
  else if (sum(x==1, na.rm=TRUE)==3) replace( x, x==1, 1/3) 
  else x}))

new.df3 <- t(apply(df3, 2, FUN = function(x) {
  if(sum(x==1, na.rm=TRUE) ==2)  replace(x, x==1, 0.5)
  else if (sum(x==1, na.rm=TRUE)==3) replace( x, x==1, 1/3) 
  else x}))

I am able to create what I want with brute force as in the above example, but there must be a better (more concise) way. I'd greatly appreciate some help.

lapply dataframe r loops
3 Answers

Store your code in a function, store your dataframes in a list and then use lapply to loop over each list element with your function. 

 recalc <- function(df) {
        t(apply(df, 2, FUN = function(x) {
            if(sum(x==1, na.rm=TRUE) ==2)  replace(x, x==1, 0.5)
            else if (sum(x==1, na.rm=TRUE)==3) replace( x, x==1, 1/3) 
            else x}))
    } 


lapply(dflist, function(df) recalc(df))

[[1]]
   [,1]      [,2] [,3]      [,4]      [,5] [,6]
v1    0 0.3333333  0.0 0.3333333 0.3333333    0
v2    0 0.0000000  1.0 0.0000000 0.0000000    0
v3    0 0.0000000  0.5 0.5000000 0.0000000    0

[[2]]
     [,1]      [,2] [,3]      [,4]      [,5] [,6]
v3      0 0.0000000  0.5 0.5000000 0.0000000    0
v3.1    0 0.0000000  0.5 0.5000000 0.0000000    0
v1      0 0.3333333  0.0 0.3333333 0.3333333    0

[[3]]
     [,1]      [,2] [,3]      [,4]      [,5] [,6]
v1      0 0.3333333  0.0 0.3333333 0.3333333    0
v3      0 0.0000000  0.5 0.5000000 0.0000000    0
v1.1    0 0.3333333  0.0 0.3333333 0.3333333    0
user_114173140

Similar to the other answer, but a little more modular and a improved version of your function:

## Put your data frames in a list
# df_list = list(df1, df2, df3)
df_list = mget(ls(pattern = "df[0-9]"))

## Write a function to modify one column
replace_ones = function(x) {
  sx = sum(x == 1, na.rm = TRUE)
  if(sx > 1) {
    x = replace(x, x == 1, 1 / sx)
  }
  return(x)
}

## Wrap it to modify a data frame:
replace_ones_df = function(df) {
  df[] = lapply(df, replace_ones)
  return(df)
}

## Apply the function to all columns of all data frames:
result_list = lapply(df_list, replace_ones_df)
# $df1
#          v1 v2  v3
# 1 0.0000000  0 0.0
# 2 0.3333333  0 0.0
# 3 0.0000000  1 0.5
# 4 0.3333333  0 0.5
# 5 0.3333333  0 0.0
# 6 0.0000000  0 0.0
# 
# $df2
#    v3 v3.1        v1
# 1 0.0  0.0 0.0000000
# 2 0.0  0.0 0.3333333
# 3 0.5  0.5 0.0000000
# 4 0.5  0.5 0.3333333
# 5 0.0  0.0 0.3333333
# 6 0.0  0.0 0.0000000
# 
# $df3
#          v1  v3      v1.1
# 1 0.0000000 0.0 0.0000000
# 2 0.3333333 0.0 0.3333333
# 3 0.0000000 0.5 0.0000000
# 4 0.3333333 0.5 0.3333333
# 5 0.3333333 0.0 0.3333333
# 6 0.0000000 0.0 0.0000000
user_9030610

Instead of manually counting the number of 1s in the binary column and using if/else, divide the datasets placed in a list with column sums (colSums)

lapply(mget(paste0("df", 1:3)), function(x) x/colSums(x)[col(x)])
#$df1
#         v1 v2  v3
#1 0.0000000  0 0.0
#2 0.3333333  0 0.0
#3 0.0000000  1 0.5
#4 0.3333333  0 0.5
#5 0.3333333  0 0.0
#6 0.0000000  0 0.0

#$df2
#   v3 v3.1        v1
#1 0.0  0.0 0.0000000
#2 0.0  0.0 0.3333333
#3 0.5  0.5 0.0000000
#4 0.5  0.5 0.3333333
#5 0.0  0.0 0.3333333
#6 0.0  0.0 0.0000000

#$df3
#         v1  v3      v1.1
#1 0.0000000 0.0 0.0000000
#2 0.3333333 0.0 0.3333333
#3 0.0000000 0.5 0.0000000
#4 0.3333333 0.5 0.3333333
#5 0.3333333 0.0 0.3333333
#6 0.0000000 0.0 0.0000000
user_37322710
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