I would like to populate an existing empty dataframe with file information using a list and the file.info
function. I've been doing the same task using a for
loop, but would like to learn how to use the apply
family and thought this would be a nice easy example.
我想使用list和file.info函数填充包含文件信息的现有空数据帧。我一直在使用for循环执行相同的任务,但是想学习如何使用apply系列,并认为这将是一个很好的简单示例。
My list...
listOfFiles_M <- c("I:\\temp\\APIS2//APIS01/WAV/APIS01_20170414_150000.wav", "I:\\temp\\APIS2//APIS01/WAV/APIS01_20170414_160000.wav",
"I:\\temp\\APIS2//APIS01/WAV/APIS01_20170414_170000.wav", "I:\\temp\\APIS2//APIS01/WAV/APIS01_20170414_180000.wav"
)
My empty dataframe...
我的空数据框......
m_files <- structure(list(size = numeric(0), isdir = logical(0), mode = structure(integer(0), class = "octmode"),
mtime = structure(numeric(0), class = c("POSIXct", "POSIXt"
)), ctime = structure(numeric(0), class = c("POSIXct", "POSIXt"
)), atime = structure(numeric(0), class = c("POSIXct", "POSIXt"
)), exe = character(0)), .Names = c("size", "isdir", "mode",
"mtime", "ctime", "atime", "exe"), row.names = character(0), class = "data.frame")
My function...
test.info <- function(i,x){
print (i)
x[i,]=c(file.info(i))
}
And I thought I should use lapply
thusly...
而且我认为我应该这样使用lapply ......
lapply(listOfFiles_M, test.info)
And here is an example of what I would like a populated m_files
to look like...
这里有一个例子,我希望填充的m_files看起来像......
m_files <- structure(list(rn = c("I:\\temp\\APIS2//APIS01/WAV/APIS01_20170414_150000.wav",
"I:\\temp\\APIS2//APIS01/WAV/APIS01_20170414_160000.wav", "I:\\temp\\APIS2//APIS01/WAV/APIS01_20170414_170000.wav",
"I:\\temp\\APIS2//APIS01/WAV/APIS01_20170414_180000.wav"), size = c(9601276,
9601276, 9601276, 9601276), isdir = c(FALSE, FALSE, FALSE, FALSE
), mode = structure(c(438L, 438L, 438L, 438L), class = "octmode"),
mtime = structure(c(1492200300, 1492203900, 1492207500, 1492211100
), class = c("POSIXct", "POSIXt")), ctime = structure(c(1537974713.78911,
1537974713.85152, 1537974713.89832, 1537974713.92952), class = c("POSIXct",
"POSIXt")), atime = structure(c(1537974713.78911, 1537974713.85152,
1537974713.89832, 1537974713.92952), class = c("POSIXct",
"POSIXt")), exe = c("no", "no", "no", "no")), .Names = c("rn",
"size", "isdir", "mode", "mtime", "ctime", "atime", "exe"), row.names = c(NA,
-4L), class = "data.frame")
EDIT: I should have also mentioned that there is a large list, ~200,000 items, so rbind
is probably not a good solution.
编辑:我应该也提到有一个大的列表,约200,000项,所以rbind可能不是一个好的解决方案。
2 个解决方案
#1
1
Simply pass your list of files into file.info
which can receive more than 1 value as input and returns a data frame as according to docs, ?file.info
.
只需将您的文件列表传递到file.info,该文件可以接收多于1个值作为输入,并根据文档“?file.info”返回数据框。
final_df <- file.info(listOfFiles_M)
No need to initialize an empty data frame and map values to it or rbind
iterative returned objects.
无需初始化空数据框并将值映射到它或rbind迭代返回的对象。
#2
0
I assume the function file.info
is designed to take a name of a file and then spit out a vector of length 7 which you use to populate a row.
我假设函数file.info被设计为获取文件的名称,然后吐出长度为7的向量,用于填充行。
Just a recommendation, this is a bit hard to test when we do not have file.info
function's output for at least 1 file. So I would recommend simplifying your m_files data frame when you post.
只是一个建议,当我们没有至少1个文件的file.info函数输出时,这有点难以测试。因此,我建议您在发布时简化m_files数据框。
I believe the only issue is that you need to specify the x argument in your lapply.
我相信唯一的问题是你需要在你的lapply中指定x参数。
lapply(listOfFiles_M, test.info, x = m_files)
the ...
argument in apply is for you to list other arugments the function you pass apply may need, in this case it is test.info
.
apply中的...参数是为您列出您传递的函数可能需要的其他语句,在这种情况下它是test.info。
#1
1
Simply pass your list of files into file.info
which can receive more than 1 value as input and returns a data frame as according to docs, ?file.info
.
只需将您的文件列表传递到file.info,该文件可以接收多于1个值作为输入,并根据文档“?file.info”返回数据框。
final_df <- file.info(listOfFiles_M)
No need to initialize an empty data frame and map values to it or rbind
iterative returned objects.
无需初始化空数据框并将值映射到它或rbind迭代返回的对象。
#2
0
I assume the function file.info
is designed to take a name of a file and then spit out a vector of length 7 which you use to populate a row.
我假设函数file.info被设计为获取文件的名称,然后吐出长度为7的向量,用于填充行。
Just a recommendation, this is a bit hard to test when we do not have file.info
function's output for at least 1 file. So I would recommend simplifying your m_files data frame when you post.
只是一个建议,当我们没有至少1个文件的file.info函数输出时,这有点难以测试。因此,我建议您在发布时简化m_files数据框。
I believe the only issue is that you need to specify the x argument in your lapply.
我相信唯一的问题是你需要在你的lapply中指定x参数。
lapply(listOfFiles_M, test.info, x = m_files)
the ...
argument in apply is for you to list other arugments the function you pass apply may need, in this case it is test.info
.
apply中的...参数是为您列出您传递的函数可能需要的其他语句,在这种情况下它是test.info。