rename

Iterate through two lists, check for matches and then rename

China☆狼群 提交于 2019-12-11 06:45:36
问题 Ok, so I have two lists; one is a list of song titles, the other is a list of files that is generated by running os.listdir() , which will be song mp3 files. UPDATED songs = ['The Prediction', 'Life We Chose', 'Nastradamus', 'Some of Us Have Angels', 'Project Windows', 'Come Get Me', "Shoot 'em Up", 'Last Words', 'Family', 'God Love Us', 'Quiet Niggas', 'Big Girl', 'New World', 'You Owe Me', 'The Outcome'] Each song is unicode filenames = ['Nas - Big Girl.mp3', 'Nas - Come Get Me.mp3', 'Nas -

PowerShell/Batch script to rename each file to include original name + folder name

柔情痞子 提交于 2019-12-11 06:18:56
问题 I have the following folder structure: April reports ├─01-04-2018 │ ├─approved123.pdf │ ├─approved123_V2.pdf │ └─unapproved123.pdf │ ├─02-04-2018 │ ├─approved123.pdf │ └─unapproved123.pdf ╎ ╎ └─30-04-2018 ├─approved123.pdf └─unapproved123.pdf Each folder for each day in April contains an approved and an unapproved report with the same name ("approved123" or "unapproved123"). Some contain a V2. I want to rename each one so that the "123" is removed and the folder name (the date) is included in

Rename image files on server directory

回眸只為那壹抹淺笑 提交于 2019-12-11 06:13:16
问题 I need some help in renaming some images in a directory located at /images/graphicsLib/. All image names in /graphicsLib/ have naming convention that looks like this: 400-60947.jpg. We call the "400" part of the file the prefix and we call the "60957" part the suffix. The entire file name we call the sku. So if you saw the contents of /graphicLib/ it would look like: 400-60957.jpg 400-60960.jpg 400-60967.jpg 400-60968.jpg 402-60988.jpg 402-60700.jpg 500-60725.jpg 500-60733.jpg etc... Using C#

Change all R columns names using a reference file

此生再无相见时 提交于 2019-12-11 06:05:47
问题 I am trying to rename columns in a dataframe in R . However, the renaming has circular referencing. I would like a solution to this problem, the circular referencing cannot be avoided. One way to think was to rename a column and move it to a new dataframe , hence, avoiding the circular referencing. However, I am unable to do so. The renaming reference is as follows: The current function I am using is as follows: standard_mapping <- function(mapping.col, current_name, standard_name, data){ for

myLocationEnabled changing to isMyLocationEnabled on swift when using GoogleMaps

 ̄綄美尐妖づ 提交于 2019-12-11 05:23:22
问题 Im trying to get a users current location to show with google maps API. I have been struggling with this issue for weeks and the code runs but doesn't update to current location when I change the location in the simulator. My code is below and after looking how other people are doing it they all have mapView.myLocationEnabled = true whereas for me this produces an error saying 'myLocationEnabled has been renamed to 'isMylocationEnabled'. I've seen some up to date posts (from 2 months ago) and

Batch: find file and rename it / find file and delete it

北慕城南 提交于 2019-12-11 04:59:42
问题 So... yes, Im pretty newbie. Still trying to explain as good as possible. But needing help with such thing as finding exact file from C: drive (with batch), and rename it. Also wanted to know how to find file from C: and delete it. (2 problems in 1 thread...) Problem 1. example: I want to find file called "text1.txt" from C: drive with batch, if successfully found, rename it as "text2.txt". Problem 2. example: I want to find file called "image1.jpg" from C: drive with batch, if successfully

Renaming LINQ 2 SQL Entity Properties Through Partial Classes

依然范特西╮ 提交于 2019-12-11 04:37:34
问题 Can I use partial classes to create properties that points to an association property generated by the L2S designer. Also, will I be able to use the new property in queries? How can I achieve this? 回答1: Yes you can, but you have to apply the same attributes as the linq2sql generated property i.e. [Association(Name="Test_TestData", Storage="_TestDatas", ThisKey="SomeId", OtherKey="OtherId")] public System.Data.Linq.EntitySet<TestData> MyTestDatas { get { return this.TestDatas; } } TestDatas

Renaming Folder Structure in Batch

北城余情 提交于 2019-12-11 04:34:53
问题 ) In order to make the company (where I'm working) operate more organized and efficiently we've created an approved folder structure for new projects. New projects and Live projects reside on a file server and are at the same level as each other. (Live projects have made it through planning, new projects sometimes only go as far as planning) The project folder structure can be a meticulous to create each time we start a new project, therefore I made a windows batch file to create it. You can

Removing Middle of Filename

我们两清 提交于 2019-12-11 04:30:41
问题 I wish to remove the middle of filenames that look like the following: Wm-no-Bis.GAGATTCC-ATAGAGGC.L001.R1.001.fastq.trim.R2.fastq Pek-no-Bis.GAGATTCC-AGGCGAAG.L001.R1.001.fastq.trim.R1.fastq To Get Wm-no-Bis.trim.R2.fastq Pek-no-Bis.trim.R1.fastq As you can see the beginning and end of the files are slightly variable. I have attempted to use rename as follows: rename 's/.*.*.*.*.*.trim.//g' Wm-no-Bis.GAGATTCC-ATAGAGGC.L001.R1.001.fastq.trim.R2copy.fastq This removes the entire file name

list - rename specific data.frames column with lapply

会有一股神秘感。 提交于 2019-12-11 04:24:53
问题 I have got a list with 10 data.frames and I need to rename ONLY one column of each data.frame. The column to rename is the no. 7 and I think I can do the trick with lapply. Here what I tried without success: lst <- lapply(lst, function(x) colnames(x)[7] <- 'new_name') I think I am really close to the solution but obviously I am missing something. Thanks 回答1: You need to use {} and return x : lst <- lapply(lst, function(x) {colnames(x)[7] <- 'new_name'; x}) Or lst <- lapply(lst, function(x) {