子集数据以仅包含名称与条件匹配的列 [英] Subset data to contain only columns whose names match a condition

问题描述

有没有办法根据以特定字符串开头的列名对数据进行子集化?我有一些列类似于 ABC_1 ABC_2 ABC_3 和一些类似于 XYZ_1, XYZ_2,XYZ_3 假设.

Is there a way for me to subset data based on column names starting with a particular string? I have some columns which are like ABC_1 ABC_2 ABC_3 and some like XYZ_1, XYZ_2,XYZ_3 let's say.

如何仅根据包含上述文本部分的列(比方说，ABC 或 XYZ)来子集我的 df?我可以使用索引，但列在数据中过于分散，而且硬编码过多.

How can I subset my df based only on columns containing the above portions of text (lets say, ABC or XYZ)? I can use indices, but the columns are too scattered in data and it becomes too much of hard coding.

另外，我只想包含这些列中的每一列的行，其中任何值都是 >0 所以如果上面的 6 列有一个 1 在行中，它会切入我的最终数据框.

Also, I want to only include rows from each of these columns where any of their value is >0 so if either of the 6 columns above has a 1 in the row, it makes a cut into my final data frame.

推荐答案

在您的 data.frame 的名称上尝试 grepl.grepl 将正则表达式与目标匹配，如果找到匹配则返回 TRUE，否则返回 FALSE.该函数是矢量化的，因此您可以传递字符串向量进行匹配，您将获得返回的布尔值向量.

Try grepl on the names of your data.frame. grepl matches a regular expression to a target and returns TRUE if a match is found and FALSE otherwise. The function is vectorised so you can pass a vector of strings to match and you will get a vector of boolean values returned.

#  Data
df <- data.frame( ABC_1 = runif(3),
            ABC_2 = runif(3),
            XYZ_1 = runif(3),
            XYZ_2 = runif(3) )

#      ABC_1     ABC_2     XYZ_1     XYZ_2
#1 0.3792645 0.3614199 0.9793573 0.7139381
#2 0.1313246 0.9746691 0.7276705 0.0126057
#3 0.7282680 0.6518444 0.9531389 0.9673290

#  Use grepl
df[ , grepl( "ABC" , names( df ) ) ]
#      ABC_1     ABC_2
#1 0.3792645 0.3614199
#2 0.1313246 0.9746691
#3 0.7282680 0.6518444

#  grepl returns logical vector like this which is what we use to subset columns
grepl( "ABC" , names( df ) )
#[1]  TRUE  TRUE FALSE FALSE

要回答第二部分，我会制作子集 data.frame，然后制作一个向量来索引要保留的行(逻辑向量)，就像这样...

To answer the second part, I'd make the subset data.frame and then make a vector that indexes the rows to keep (a logical vector) like this...

set.seed(1)
df <- data.frame( ABC_1 = sample(0:1,3,repl = TRUE),
            ABC_2 = sample(0:1,3,repl = TRUE),
            XYZ_1 = sample(0:1,3,repl = TRUE),
            XYZ_2 = sample(0:1,3,repl = TRUE) )

# We will want to discard the second row because 'all' ABC values are 0:
#  ABC_1 ABC_2 XYZ_1 XYZ_2
#1     0     1     1     0
#2     0     0     1     0
#3     1     1     1     0


df1 <- df[ , grepl( "ABC" , names( df ) ) ]

ind <- apply( df1 , 1 , function(x) any( x > 0 ) )

df1[ ind , ]
#  ABC_1 ABC_2
#1     0     1
#3     1     1

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