计算相似度百分比或计算两个以上对象之间的相关性 [英] Compute similarity percentage OR Compute correlation between more than 2 objects

问题描述

考虑一下，我有四个对象(a,b,c,d)，我请五个人根据其外观或其他东西来标记它们(类别1或2).五个人为这些物体提供的标签显示为

Consider I have four objects (a,b,c,d), and I ask five persons to label them (category 1 or 2) according to their physical appearance or something else. The labels provided by five persons for these objects are shown as

df <- data.frame(a = c(1,2,1,2,1), b=c(1,2,2,1,1), c= c(2,1,2,2,2), d=c(1,2,1,2,1))

以表格格式，

 ---------
  a b c d
 ---------
  1 1 2 1
  2 2 1 2
  1 2 2 1
  2 1 2 2
  1 1 2 1
----------

现在，我想计算一组对象被赋予相同标签(1或2)的次数的百分比.例如，对象a，b和d在5个人中有3个人被赋予了相同的标签.因此其百分比为3/5(= 60％).由于对象a和d被所有人赋予相同的标签，因此其百分比为5/5(= 100％)

Now I want to calculate the percentage of times a group of objects were given the same label (either 1 or 2). For example, objects a, b and d were given the same label by 3 persons out of 5 persons. So its percentage is 3/5 (=60%). While as objects a and d were given same labels by all the people, so its percentage is 5/5 (=100%)

我可以手动计算该统计量，但是在我的原始数据集中，我有50个这样的对象，人是30，标签是4(1、2、3和4).如何自动为更大的数据集计算此类统计信息? R中是否有任何可以计算此类统计信息的软件包/工具?

I can calculate this statistic manually, but in my original dataset, I have 50 such objects and the people are 30 and the labels are 4 (1,2,3, and 4). How can I compute such statistics for this bigger dataset automatically? Are there any existing packages/tools in R which can calculate such statistics?

注意:一个组可以是任意大小.在第一个示例中，组由a，b和d组成，而在第二个示例中，组由a和d组成.

Note: A group can be of any size. In the first example, a group consists of a,b and d while as second example group consists of a and d.

推荐答案

此处有两个任务:首先，列出所有相关组合的列表，其次，评估和汇总行相似性. combn可以启动第一个任务，但是需要一点按摩才能将结果整理到一个整齐的列表中.第二个任务可以用prop.table处理，但是在这里直接计算更简单.

There are two tasks here: firstly, making a list of all the relevant combinations, and secondly, evaluating and aggregating rowwise similarity. combn can start the first task, but it takes a little massaging to arrange the results into a neat list. The second task could be handled with prop.table, but here it's simpler to calculate directly.

在这里，我使用了tidyverse语法(主要是purrr，这对处理列表很有帮助)，但是如果您愿意，可以转换为基数.

Here I've used tidyverse grammar (primarily purrr, which is helpful for handling lists), but convert into base if you like.

library(tidyverse)

map(2:length(df), ~combn(names(df), .x, simplify = FALSE)) %>%    # get combinations
    flatten() %>%    # eliminate nesting
    set_names(map_chr(., paste0, collapse = '')) %>%    # add useful names
    # subset df with combination, see if each row has only one unique value
    map(~apply(df[.x], 1, function(x){n_distinct(x) == 1})) %>% 
    map_dbl(~sum(.x) / length(.x))    # calculate TRUE proportion

##   ab   ac   ad   bc   bd   cd  abc  abd  acd  bcd abcd 
##  0.6  0.2  1.0  0.2  0.6  0.2  0.0  0.6  0.2  0.0  0.0 

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