如何计算以R开头的数据帧中的单元格百分比？ [英] How to calculate percentage of cells in data frame that start with sequence in R?

问题描述

我有数据如下：

 行1行2行3行4行5行6行7 
 abc89 abc62 abc513 abc512 abc81 abc10 
 abc6 pop abc11 abc4 big 13 abc15 
 abc90 abc16 abc123 abc33 abc22 abc08 9 
 111 abc15 abc72 abc36 abc57 abc9 abc55 
  

我想计算以abc开头的数据框中的单元格百分比。例如：这里总共有28个单元格。这可以通过 prod（dim（df））获得。所以我需要以abc开头的单元格数，然后用 prod（dim（df））分隔。这里的答案是0.785。如何在R中完成？

解决方案

我会使用：

 > mean（grepl（^ abc，unlist（dat）））
 [1] 0.7857143 
  

使用意味着意味着您不必单独获取分子和分母。 grepl 是 grep 的逻辑版本 - 它返回 TRUE 每当^ abc（即，以 abc 开头的字符串）。 回想一下，伯努利向量的平均值是成功的百分比。 / p>

如果您想按行或列进行排列，您将使用应用，例如 apply（dat，1，function（x）mean（grepl（^ abc，x）））获得行方式。

I have data that looks like:

Row 1     Row 2     Row 3     Row 4     Row 5     Row 6     Row7
abc89     abc62     67        abc513    abc512    abc81     abc10
abc6      pop       abc11     abc4      giant     13        abc15
abc90     abc16     abc123    abc33     abc22     abc08     9
111       abc15     abc72     abc36     abc57     abc9      abc55

I would like to calculate the percentage of cells in the data frame that begin with "abc". For example: there are 28 total cells here. This can be gotten by prod(dim(df)). So I need the # of cells that start with "abc" and then divide it by prod(dim(df)). Here the answer would be 0.785. how can this be done in R?

解决方案

I would use:

> mean(grepl("^abc",unlist(dat)))
[1] 0.7857143

Using mean means you don't have to get the numerator and denominator yourself separately. grepl is the logical version of grep--it returns TRUE whenever "^abc" (i.e., a string beginning with abc) is found. Recall that the average of a Bernoulli vector is the percentage of successes.

If you wanted to do this by row or by column you'd use apply, e.g. apply(dat,1,function(x)mean(grepl("^abc",x))) to get the row-wise means.

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