r - 如何在R中的数据框的同一列中显示项目之间的共享特征

Question

如果数据框包含项目列、它们的特征及其特征值，如何在数据框的同一列中显示项目之间的共享特征？

这是输入数据框

df <- data.frame(group = c(rep(c('A', 'B', 'C'), 3)),
                                   feature = c('x','x','x','y','y','z','z','w','t'),
                                   value=c(1,2,1,3,2,1,2,2,3))

这是所需输出的示例，它比较了任意两对 A&B、A&C、B&C 和过滤器的共同特征:

df_desired <- data.frame(group1 =c('A','A','B'), group2 = c('B','C','C'), shared_feature = c('x','x','x'), value1 = c(1, 1,2), value2 = c(1, 2,1))

Answer 1

您可以尝试以下操作:

# Added stringsAsFactors=F argument
df <- data.frame(group = c(rep(c('A', 'B', 'C'), 3)),
                 feature = c('x','x','x','y','y','z','z','w','t'),
                 value=c(1,2,1,3,2,1,2,2,3),stringsAsFactors = F)

df_desired <- data.frame(group1 =c('A','A','B'), group2 = c('B','C','C'), shared_feature = c('x','x','x'), value1 = c(1, 1,2), value2 = c(1, 2,1))

# For rbindlist function
library(data.table)

# Keep only features that are available for every group
df_agg = aggregate(value ~ feature, data = df, FUN = length)
shared_feats = df_agg$feature[df_agg$value==length(unique(df$group))]
df = df[df$feature %in% shared_feats,]

# A function that takes a feat_df containing the values of one feature for each group,
# and converts it to our expected output.
create_comb_df <- function(feat_df)
{
  df2 = as.data.frame(t(combn(feat_df$group,2)))
  colnames(df2) = c('group1','group2')
  df2$feature = feat_df$feature[1]
  df2$value1 = feat_df$value[match(df2$group1,feat_df$group)]
  df2$value2 = feat_df$value[match(df2$group2,feat_df$group)]
  return(df2)
}

# Create final output
rbindlist(lapply(split(df,as.character(df$feature)),create_comb_df))

输出:

   group1 group2 feature value1 value2
1:      A      B       x      1      2
2:      A      C       x      1      1
3:      B      C       x      2      1

或者，要获取所有共享功能，请替换

shared_feats = df_agg$feature[df_agg$value==length(unique(df$group))]

和

shared_feats = df_agg$feature[df_agg$value>1]

结果是:

   group1 group2 feature value1 value2
1:      A      B       x      1      2
2:      A      C       x      1      1
3:      B      C       x      2      1
4:      A      B       y      3      2
5:      C      A       z      1      2

希望这可以帮助!