如何使用lapply并将其粘贴到列表中的多个数据框上

问题描述

我不能结合使用lapply和paste来组合两个列，以用于列表中包含的多个数据框.我已经查看了多个来源，但找不到答案.

I can't combine the use of lapply and paste to combine two columns for multiple dataframes contained within a list. I've looked through multiple sources, but can't find the answer.

此答案应用粘贴在用于获取字符串列表的向量是关于组合列表中的行，而不是组合列以获得向量.

This answerApply paste over a list of vectors to get a list of strings is about combining rows within lists, not combining columns to obtain a vector.

此答案说明了如何选择列而不将它们粘贴在一起在上面使用lapply数据框列表

This answer explains how to select columns but not paste them together using lapply on a list of dataframes

此页面说明了如何访问列表中的列，但没有说明如何将它们组合在一起 https://www.datacamp.com/community/tutorials/r-tutorial-apply-family

This page explains how to access columns in lists, but not how to combine them together https://www.datacamp.com/community/tutorials/r-tutorial-apply-family

这个答案是最接近的，但是我不能对文本

This answer is the closest, but I can't use transform on text Performing an operation on multiple columns in a list of data

下面的代码是一个示例，该示例成功创建了输出，但未使用lapply.我将在许多不同的数据帧上应用此功能，因此需要使用lapply.大概我可以使用for for循环，但是lapply可能更干净.我需要使用lapply而不是sapply，因为数据帧的长度将不同.

The code below is an example which successfully creates the output, but does not use lapply. I will be applying this function over many different dataframes, so need to use lapply. Presumably I could use a for i loop, but lapply is probably cleaner. I need to use lapply, rather than sapply, as the dataframes will have different lengths.

我想用一个lapply函数替换开始output1，output2和output的三行，该函数将在此示例中跨两个数据帧工作.我认为应该是

I want to replace the three lines that start output1, output2 and output with an lapply function that will work across the two dataframes in this example. I think it should be something like

output <-lapply(mylist, paste($Artist,$Song))

但这不起作用.预先感谢.

But that doesn't work. Thanks in advance.

Artist <- c("Drake", "Ed Sheeran", "Bruno Mars", "Camilla Cabello", "BlocBoy")    
Song <- c("Gods Plan", "Perfect", "Finesse", "Havana", "Look Alive")
Current <- data.frame(cbind(Artist,Song))
Artist <- c("Gucci Mane", "Migos", "Daft Punk", "Chainsmokers")
Song <- c("Black Beatles", "Bad and Boujee", "Starboy", "Closer")
Past <- data.frame(cbind(Artist,Song))

mylist <- list(Current,Past)

output1  <- paste(mylist[[1]]$Artist,mylist[[1]]$Song)
output2  <- paste(mylist[[2]]$Artist,mylist[[2]]$Song)
output <-list(output1,output2)
print(output)

推荐答案

我认为这是您要使用的:

I think this is what you're going for:

lapply(mylist, function(x) paste(x$Artist, x$Song, sep=" "))

我们只需要定义小的匿名函数来连接列即可.

We just need to define the little anonymous function to concatenate the columns.

[[1]]
[1] "Drake Gods Plan"        "Ed Sheeran Perfect"     "Bruno Mars Finesse"    
[4] "Camilla Cabello Havana" "BlocBoy Look Alive"    

[[2]]
[1] "Gucci Mane Black Beatles" "Migos Bad and Boujee"    
[3] "Daft Punk Starboy"        "Chainsmokers Closer" 

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