使用R函数lapply和read.sql.csv [英] Using R functions lapply and read.sql.csv

问题描述

我尝试使用如下列表打开多个csv文件：

 文件名<  -  list.files （temp，pattern =*。csv，full.names = TRUE）
  

发现使用lapply和read.csv打开temp目录中的所有文件的示例，但我知道appriori什么数据我需要从文件中提取，所以为了节省时间阅读我想使用这个SQL扩展;

  somefile = read.csv.sql（temp / somefile.csv，sql =select * from file，eol = \ n）
  

但是我无法将这两个功能合并到一个命令我可以读取应用相同sql查询的目录中的所有文件。

有没有人成功了？

  files<  -  list.files（。，pattern =*。csv）
 df.list<  -  sapply（filenames，read.csv.sql，sql =select * from file，eol =\\\
，simplify = F）
  

如果你想要他们所有组合：

  df < -  ldply（filenames，read.csv.sql，sql =select *从文件，eol =\\\
）
  

I am trying to open multiple csv files using a list such as the below;

filenames <- list.files("temp", pattern="*.csv", full.names=TRUE)

I have found examples that use lapply and read.csv to open all the files in the temp directory, but I know appriori what data i need to extract from the file, so to save time reading i want to use the SQL extension of this;

somefile = read.csv.sql("temp/somefile.csv", sql="select * from file ",eol="\n")

However i am having trouble combining these two pieces of functionality into a single command such that i can read all the files in a directory applying the same sql query.

Has anybody had success doing this?

解决方案

If you want a list of dataframes from each file (assuming your working directory contains the .csv files):

files <- list.files(".", pattern="*.csv")
df.list <- sapply(filenames, read.csv.sql,sql="select * from file ",eol="\n",simplify=F)

Or if you want them all combined:

df <- ldply(filenames, read.csv.sql,sql="select * from file ",eol="\n")

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