transform columns of a list into a new dataframe, if the list is equal to zero add nas in the new columns in r

You can achieve this by using the ifelse function along with mutate from the dplyr package in R. Here is a sample code snippet that does this:

main.r
# Load necessary libraries
library(dplyr)

# Create a sample dataframe
df <- data.frame(
  id = c(1, 2, 3),
  values = list(c(10, 20), 0, c(30, 40))
)

# Transform columns of a list into a new dataframe
df_new <- df %>%
  mutate(
    col1 = ifelse(lengths(values) == 0, NA, sapply(values, function(x) x[1])),
    col2 = ifelse(lengths(values) == 0 | lengths(values) < 2, NA, sapply(values, function(x) x[2]))
  )

# Print the new dataframe
print(df_new)
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19 lines

In this code, lengths(values) is used to check the length of each list in the values column. If the length is 0 (i.e., the list is empty), it assigns NA to the new columns. Otherwise, it uses sapply to extract the first and second elements of each list.

If you have a list with more than two elements and you want to create new columns for all of them, you can use a loop to generate the new columns dynamically:

main.r
# Load necessary libraries
library(dplyr)
library(purrr)

# Create a sample dataframe
df <- data.frame(
  id = c(1, 2, 3),
  values = list(c(10, 20), 0, c(30, 40, 50))
)

# Find the maximum length of the lists
max_length <- max(map_int(df$values, length))

# Transform columns of a list into a new dataframe
for (i in seq_len(max_length)) {
  col_name <- paste0("col", i)
  df[[col_name]] <- ifelse(map_int(df$values, length) < i, NA, sapply(df$values, function(x) x[i]))
}

# Print the new dataframe
print(df)
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22 lines

This code first finds the maximum length of the lists in the values column. Then it uses a loop to generate new columns for each position up to the maximum length. If a list is shorter than the current position, it assigns NA to the corresponding new column. Otherwise, it extracts the value at the current position from each list.

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