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  2. Get a list from Pandas DataFrame column headers

    stackoverflow.com/questions/19482970

    Create a list of keys/columns - object method to_list() and the Pythonic way: my_dataframe.keys().to_list() list(my_dataframe.keys()) Basic iteration on a DataFrame returns column labels: [column for column in my_dataframe] Do not convert a DataFrame into a list, just to get the column labels.

  3. Naming list elements in R - Stack Overflow

    stackoverflow.com/questions/38643000

    However, the reason that your second approach (names(filList[i]) <- ) does not work is NOT that "you create a unbound temporary variable whose names you change but which never matters since its just a temporary variable" but rather that the default sub-setting assignment operator [<-does not transfer names (which makes sense since names are ...

  4. Access and preserve list names in lapply function

    stackoverflow.com/questions/9469504

    Building on joran's answer, and precising it: The sapply(USE.NAMES=T) wrapper will indeed set as names of the final result the values of the vector you are iterating over (and not its names attribute like lapply), but only if these are characters.

  5. names(LIST) is not correct because it returns "A" and "B". Just for context I am plotting a series of data frames that are stored in a list. As I come to each data.frame I want to include the name of the data.frame as the title.

  6. The output (which includes the relative path, not just the file name) follows this format: dir 1/dir 2/filename.ext. dir 3/dir 4/other filename.ext. You can also do, for example. git diff --name-only HEAD~10 HEAD~5. to see the differences between the tenth latest commit and the fifth latest (or so).

  7. Return a possibly-empty list of path names that match pathname, which must be a string containing a path specification. pathname can be either absolute (like /usr/src/Python-1.5/Makefile) or relative (like ../../Tools/*/*.gif), and can contain shell-style wildcards. Broken symlinks are included in the results (as in the shell)....

  8. Either way, you could use os.walk to do this: os.walk(directory) will yield a tuple for each subdirectory. Ths first entry in the 3-tuple is a directory name, so. [x[0] for x in os.walk(directory)] should give you all of the subdirectories, recursively. Note that the second entry in the tuple is the list of child directories of the entry in the ...

  9. 5. The below post gives the solution for your scenario. **dir /s /b /o:gn**. /S Displays files in specified directories and all subdirectories. /B Uses bare format (no heading information or summary). /O List by files in sorted order. :gn, g sorts by folders and then files, and n puts those files in alphabetical order.

  10. In a query editor, if you highlight the text of table name (ex dbo.MyTable) and hit ALT+F1, you'll get a list of column names, type, length, etc. ALT+F1 while you've highlighted dbo.MyTable is the equivalent of running EXEC sp_help 'dbo.MyTable' according to this site

  11. Batch File; List files in directory, only filenames?

    stackoverflow.com/questions/23228983

    4.Save "list.bat" Thats it. now you can copy & paste this "list.bat" file any of your folder location and double click it, it will create a "fileslist.txt" along with that directory folder and file name list. Sample Output: Note: If you want create file name list along with sub folder, then you can create batch file with bellow code.