python - 在 R 中创建我的数据框列的所有可能组合

Question

我正在使用 R 中的数据框，其中包含以下列:

我想创建一个新的数据框，其中包含我的列的所有可能组合(数据应该相乘)，并按照下表填充数据。棘手的部分是，如果变量的较高组合的值为 1，则所有较低的组合都应设为零。例如，在第三行； a=1，b=1，c=1。此处，abc = 1 并且所有其他组合都应设为 NA，因为它们是 abc 的子集。我的原始数据框中有 6 列，因此使事情变得更加复杂。

Answer 1

我建议使用

na.act <- getOption('na.action')
options('na.action' = na.pass)

# Create wide data.
df_wide <- as.data.frame(model.matrix(~ (.)^3 - RESPID,  #replace 3 with the number of columns in your data
                        data = df))
options('na.action' = na.act)
df_wide$RESPID <- df$RESPID

编辑:

花了一点时间，我想出了一个解决方案。它需要大量转换数据、分组、嵌套，并且可能比@Jonas 提供的答案效率更低。但它是逐步完成工作的。它是……有点可读性，并在其他方面进行了评论。它是使用 3 个基本步骤作为我的主要想法完成的。它们在代码中有描述，但代码本身没有很好的记录。

初始化:

df <- data.frame(RESPID = 1:3, A = c(1, NA, 1), B = c(NA, 1, 1), C = c(1, 1, 1))

# Start by creating a wide data.frame using model.matrix
na.act <- getOption('na.action')
options('na.action' = na.pass)
df_wide <- as.data.frame(model.matrix(~ (.- RESPID)^3 - 1,  #replace 3 with the number of columns in your data
                                      data = df))
options('na.action' = na.act)
df_wide$RESPID <- df$RESPID
df_wide

第一步:

# Next lets get the groups that should actually be filled.
# To do this, we'll 
# first) make a "long" format, with all the values that are currently filled.
# second) Find the active columns (A, B, C in this case) for the values that are filled, and count the number of active columns
# third) Find the maximum number of active columns for each active column (A, B, C)
# It is not very readable.
library(tidyverse)
library(stringr)
# Create a long version of the data, and split columns into multiple names.
df_longer <- pivot_longer(df_wide, cols = 1:7) %>% 
  mutate(name_split = str_split(name, ':')) %>% 
  mutate(col_count = lengths(name_split)) %>%
  unnest_wider(name_split) %>%
  # Sort by the number of letters used.
  arrange(col_count) %>% 
  rename('First_active' = '...1',
         'Second_Active' = '...2',
         'Third_Active' = '...3') %>% 
  pivot_longer(cols = 4:6, 
               names_to = 'second_name',
               values_to = 'second_value') 

第 2 步(和第 3 步):

# Find the columns with maximum number of letters used:
max_indx <- 
  df_longer %>% group_by(second_name) %>%
    drop_na(second_value, value) %>% 
    ungroup() %>%
    select(RESPID, second_value, value, col_count) %>%
    group_by(second_value, RESPID) %>%
    summarize(indx_max = max(col_count), .groups = 'drop') %>% 
    group_by(RESPID) %>%
    nest(data = second_value) %>% 
    select(data) %>%
    mutate(colname = paste0(data[[1]][[1]], collapse = ':')) %>% 
    ungroup() %>% 
    select(-data)

# Print for visualization
max_indx
# A tibble: 3 x 2
  RESPID colname
   <int> <chr>  
1      1 A:C    
2      3 A:B:C  
3      2 B:C  

第 3 步(或第 4 步？):

# Now that we have the max indices, lets overwrite the values in df_wide
for(i in seq_len(nrow(max_indx))){
  name <- max_indx$colname[max_indx$RESPID == i]
  df_wide[df_wide$RESPID == i, name] <- 1
  df_wide[df_wide$RESPID == i, colnames(df_wide)[!colnames(df_wide) %in% c(name, 'RESPID')]] <- 0
}
df_wide
  A B C A:B A:C B:C A:B:C RESPID
1 0 0 0   0   1   0     0      1
2 0 0 0   0   0   1     0      2
3 0 0 0   0   0   0     1      3