How to avoid Python/Pandas creating an index in a saved csv?

| |

👻 Check our latest review to choose the best laptop for Machine Learning engineers and Deep learning tasks!

I am trying to save a csv to a folder after making some edits to the file.

Every time I use pd.to_csv("C:/Path of file.csv") the csv file has a separate column of indexes. I want to avoid printing the index to csv.

I tried:

pd.read_csv("C:/Path to file to edit.csv", index_col = False)

And to save the file...

pd.to_csv("C:/Path to save edited file.csv", index_col = False)

However, I still got the unwanted index column. How can I avoid this when I save my files?

👻 Read also: what is the best laptop for engineering students?

How to avoid Python/Pandas creating an index in a saved csv? _files: Questions

How do I list all files of a directory?

5 answers

How can I list all files of a directory in Python and add them to a list?

3467

Answer #1

os.listdir() will get you everything that"s in a directory - files and directories.

If you want just files, you could either filter this down using os.path:

from os import listdir
from os.path import isfile, join
onlyfiles = [f for f in listdir(mypath) if isfile(join(mypath, f))]

or you could use os.walk() which will yield two lists for each directory it visits - splitting into files and dirs for you. If you only want the top directory you can break the first time it yields

from os import walk

f = []
for (dirpath, dirnames, filenames) in walk(mypath):
    f.extend(filenames)
    break

or, shorter:

from os import walk

filenames = next(walk(mypath), (None, None, []))[2]  # [] if no file

3467

Answer #2

I prefer using the glob module, as it does pattern matching and expansion.

import glob
print(glob.glob("/home/adam/*"))

It does pattern matching intuitively

import glob
# All files ending with .txt
print(glob.glob("/home/adam/*.txt")) 
# All files ending with .txt with depth of 2 folder
print(glob.glob("/home/adam/*/*.txt")) 

It will return a list with the queried files:

["/home/adam/file1.txt", "/home/adam/file2.txt", .... ]

3467

Answer #3

os.listdir() - list in the current directory

With listdir in os module you get the files and the folders in the current dir

 import os
 arr = os.listdir()
 print(arr)
 
 >>> ["$RECYCLE.BIN", "work.txt", "3ebooks.txt", "documents"]

Looking in a directory

arr = os.listdir("c:\files")

glob from glob

with glob you can specify a type of file to list like this

import glob

txtfiles = []
for file in glob.glob("*.txt"):
    txtfiles.append(file)

glob in a list comprehension

mylist = [f for f in glob.glob("*.txt")]

get the full path of only files in the current directory

import os
from os import listdir
from os.path import isfile, join

cwd = os.getcwd()
onlyfiles = [os.path.join(cwd, f) for f in os.listdir(cwd) if 
os.path.isfile(os.path.join(cwd, f))]
print(onlyfiles) 

["G:\getfilesname\getfilesname.py", "G:\getfilesname\example.txt"]

Getting the full path name with os.path.abspath

You get the full path in return

 import os
 files_path = [os.path.abspath(x) for x in os.listdir()]
 print(files_path)
 
 ["F:\documentiapplications.txt", "F:\documenticollections.txt"]

Walk: going through sub directories

os.walk returns the root, the directories list and the files list, that is why I unpacked them in r, d, f in the for loop; it, then, looks for other files and directories in the subfolders of the root and so on until there are no subfolders.

import os

# Getting the current work directory (cwd)
thisdir = os.getcwd()

# r=root, d=directories, f = files
for r, d, f in os.walk(thisdir):
    for file in f:
        if file.endswith(".docx"):
            print(os.path.join(r, file))

os.listdir(): get files in the current directory (Python 2)

In Python 2, if you want the list of the files in the current directory, you have to give the argument as "." or os.getcwd() in the os.listdir method.

 import os
 arr = os.listdir(".")
 print(arr)
 
 >>> ["$RECYCLE.BIN", "work.txt", "3ebooks.txt", "documents"]

To go up in the directory tree

# Method 1
x = os.listdir("..")

# Method 2
x= os.listdir("/")

Get files: os.listdir() in a particular directory (Python 2 and 3)

 import os
 arr = os.listdir("F:\python")
 print(arr)
 
 >>> ["$RECYCLE.BIN", "work.txt", "3ebooks.txt", "documents"]

Get files of a particular subdirectory with os.listdir()

import os

x = os.listdir("./content")

os.walk(".") - current directory

 import os
 arr = next(os.walk("."))[2]
 print(arr)
 
 >>> ["5bs_Turismo1.pdf", "5bs_Turismo1.pptx", "esperienza.txt"]

next(os.walk(".")) and os.path.join("dir", "file")

 import os
 arr = []
 for d,r,f in next(os.walk("F:\_python")):
     for file in f:
         arr.append(os.path.join(r,file))

 for f in arr:
     print(files)

>>> F:\_python\dict_class.py
>>> F:\_python\programmi.txt

next(os.walk("F:\") - get the full path - list comprehension

 [os.path.join(r,file) for r,d,f in next(os.walk("F:\_python")) for file in f]
 
 >>> ["F:\_python\dict_class.py", "F:\_python\programmi.txt"]

os.walk - get full path - all files in sub dirs**

x = [os.path.join(r,file) for r,d,f in os.walk("F:\_python") for file in f]
print(x)

>>> ["F:\_python\dict.py", "F:\_python\progr.txt", "F:\_python\readl.py"]

os.listdir() - get only txt files

 arr_txt = [x for x in os.listdir() if x.endswith(".txt")]
 print(arr_txt)
 
 >>> ["work.txt", "3ebooks.txt"]

Using glob to get the full path of the files

If I should need the absolute path of the files:

from path import path
from glob import glob
x = [path(f).abspath() for f in glob("F:\*.txt")]
for f in x:
    print(f)

>>> F:acquistionline.txt
>>> F:acquisti_2018.txt
>>> F:ootstrap_jquery_ecc.txt

Using os.path.isfile to avoid directories in the list

import os.path
listOfFiles = [f for f in os.listdir() if os.path.isfile(f)]
print(listOfFiles)

>>> ["a simple game.py", "data.txt", "decorator.py"]

Using pathlib from Python 3.4

import pathlib

flist = []
for p in pathlib.Path(".").iterdir():
    if p.is_file():
        print(p)
        flist.append(p)

 >>> error.PNG
 >>> exemaker.bat
 >>> guiprova.mp3
 >>> setup.py
 >>> speak_gui2.py
 >>> thumb.PNG

With list comprehension:

flist = [p for p in pathlib.Path(".").iterdir() if p.is_file()]

Alternatively, use pathlib.Path() instead of pathlib.Path(".")

Use glob method in pathlib.Path()

import pathlib

py = pathlib.Path().glob("*.py")
for file in py:
    print(file)

>>> stack_overflow_list.py
>>> stack_overflow_list_tkinter.py

Get all and only files with os.walk

import os
x = [i[2] for i in os.walk(".")]
y=[]
for t in x:
    for f in t:
        y.append(f)
print(y)

>>> ["append_to_list.py", "data.txt", "data1.txt", "data2.txt", "data_180617", "os_walk.py", "READ2.py", "read_data.py", "somma_defaltdic.py", "substitute_words.py", "sum_data.py", "data.txt", "data1.txt", "data_180617"]

Get only files with next and walk in a directory

 import os
 x = next(os.walk("F://python"))[2]
 print(x)
 
 >>> ["calculator.bat","calculator.py"]

Get only directories with next and walk in a directory

 import os
 next(os.walk("F://python"))[1] # for the current dir use (".")
 
 >>> ["python3","others"]

Get all the subdir names with walk

for r,d,f in os.walk("F:\_python"):
    for dirs in d:
        print(dirs)

>>> .vscode
>>> pyexcel
>>> pyschool.py
>>> subtitles
>>> _metaprogramming
>>> .ipynb_checkpoints

os.scandir() from Python 3.5 and greater

import os
x = [f.name for f in os.scandir() if f.is_file()]
print(x)

>>> ["calculator.bat","calculator.py"]

# Another example with scandir (a little variation from docs.python.org)
# This one is more efficient than os.listdir.
# In this case, it shows the files only in the current directory
# where the script is executed.

import os
with os.scandir() as i:
    for entry in i:
        if entry.is_file():
            print(entry.name)

>>> ebookmaker.py
>>> error.PNG
>>> exemaker.bat
>>> guiprova.mp3
>>> setup.py
>>> speakgui4.py
>>> speak_gui2.py
>>> speak_gui3.py
>>> thumb.PNG

Examples:

Ex. 1: How many files are there in the subdirectories?

In this example, we look for the number of files that are included in all the directory and its subdirectories.

import os

def count(dir, counter=0):
    "returns number of files in dir and subdirs"
    for pack in os.walk(dir):
        for f in pack[2]:
            counter += 1
    return dir + " : " + str(counter) + "files"

print(count("F:\python"))

>>> "F:\python" : 12057 files"

Ex.2: How to copy all files from a directory to another?

A script to make order in your computer finding all files of a type (default: pptx) and copying them in a new folder.

import os
import shutil
from path import path

destination = "F:\file_copied"
# os.makedirs(destination)

def copyfile(dir, filetype="pptx", counter=0):
    "Searches for pptx (or other - pptx is the default) files and copies them"
    for pack in os.walk(dir):
        for f in pack[2]:
            if f.endswith(filetype):
                fullpath = pack[0] + "\" + f
                print(fullpath)
                shutil.copy(fullpath, destination)
                counter += 1
    if counter > 0:
        print("-" * 30)
        print("	==> Found in: `" + dir + "` : " + str(counter) + " files
")

for dir in os.listdir():
    "searches for folders that starts with `_`"
    if dir[0] == "_":
        # copyfile(dir, filetype="pdf")
        copyfile(dir, filetype="txt")


>>> _compiti18Compito Contabilità 1conti.txt
>>> _compiti18Compito Contabilità 1modula4.txt
>>> _compiti18Compito Contabilità 1moduloa4.txt
>>> ------------------------
>>> ==> Found in: `_compiti18` : 3 files

Ex. 3: How to get all the files in a txt file

In case you want to create a txt file with all the file names:

import os
mylist = ""
with open("filelist.txt", "w", encoding="utf-8") as file:
    for eachfile in os.listdir():
        mylist += eachfile + "
"
    file.write(mylist)

Example: txt with all the files of an hard drive

"""
We are going to save a txt file with all the files in your directory.
We will use the function walk()
"""

import os

# see all the methods of os
# print(*dir(os), sep=", ")
listafile = []
percorso = []
with open("lista_file.txt", "w", encoding="utf-8") as testo:
    for root, dirs, files in os.walk("D:\"):
        for file in files:
            listafile.append(file)
            percorso.append(root + "\" + file)
            testo.write(file + "
")
listafile.sort()
print("N. of files", len(listafile))
with open("lista_file_ordinata.txt", "w", encoding="utf-8") as testo_ordinato:
    for file in listafile:
        testo_ordinato.write(file + "
")

with open("percorso.txt", "w", encoding="utf-8") as file_percorso:
    for file in percorso:
        file_percorso.write(file + "
")

os.system("lista_file.txt")
os.system("lista_file_ordinata.txt")
os.system("percorso.txt")

All the file of C: in one text file

This is a shorter version of the previous code. Change the folder where to start finding the files if you need to start from another position. This code generate a 50 mb on text file on my computer with something less then 500.000 lines with files with the complete path.

import os

with open("file.txt", "w", encoding="utf-8") as filewrite:
    for r, d, f in os.walk("C:\"):
        for file in f:
            filewrite.write(f"{r + file}
")

How to write a file with all paths in a folder of a type

With this function you can create a txt file that will have the name of a type of file that you look for (ex. pngfile.txt) with all the full path of all the files of that type. It can be useful sometimes, I think.

import os

def searchfiles(extension=".ttf", folder="H:\"):
    "Create a txt file with all the file of a type"
    with open(extension[1:] + "file.txt", "w", encoding="utf-8") as filewrite:
        for r, d, f in os.walk(folder):
            for file in f:
                if file.endswith(extension):
                    filewrite.write(f"{r + file}
")

# looking for png file (fonts) in the hard disk H:
searchfiles(".png", "H:\")

>>> H:4bs_18Dolphins5.png
>>> H:4bs_18Dolphins6.png
>>> H:4bs_18Dolphins7.png
>>> H:5_18marketing htmlassetsimageslogo2.png
>>> H:7z001.png
>>> H:7z002.png

(New) Find all files and open them with tkinter GUI

I just wanted to add in this 2019 a little app to search for all files in a dir and be able to open them by doubleclicking on the name of the file in the list. enter image description here

import tkinter as tk
import os

def searchfiles(extension=".txt", folder="H:\"):
    "insert all files in the listbox"
    for r, d, f in os.walk(folder):
        for file in f:
            if file.endswith(extension):
                lb.insert(0, r + "\" + file)

def open_file():
    os.startfile(lb.get(lb.curselection()[0]))

root = tk.Tk()
root.geometry("400x400")
bt = tk.Button(root, text="Search", command=lambda:searchfiles(".png", "H:\"))
bt.pack()
lb = tk.Listbox(root)
lb.pack(fill="both", expand=1)
lb.bind("<Double-Button>", lambda x: open_file())
root.mainloop()

sep

How to print number with commas as thousands separators?

5 answers

I am trying to print an integer in Python 2.6.1 with commas as thousands separators. For example, I want to show the number 1234567 as 1,234,567. How would I go about doing this? I have seen many examples on Google, but I am looking for the simplest practical way.

It does not need to be locale-specific to decide between periods and commas. I would prefer something as simple as reasonably possible.

929

Answer #1

Locale unaware

"{:,}".format(value)  # For Python ‚â•2.7
f"{value:,}"  # For Python ‚â•3.6

Locale aware

import locale
locale.setlocale(locale.LC_ALL, "")  # Use "" for auto, or force e.g. to "en_US.UTF-8"

"{:n}".format(value)  # For Python ‚â•2.7
f"{value:n}"  # For Python ‚â•3.6

Reference

Per Format Specification Mini-Language,

The "," option signals the use of a comma for a thousands separator. For a locale aware separator, use the "n" integer presentation type instead.

929

Answer #2

I got this to work:

>>> import locale
>>> locale.setlocale(locale.LC_ALL, "en_US")
"en_US"
>>> locale.format("%d", 1255000, grouping=True)
"1,255,000"

Sure, you don"t need internationalization support, but it"s clear, concise, and uses a built-in library.

P.S. That "%d" is the usual %-style formatter. You can have only one formatter, but it can be whatever you need in terms of field width and precision settings.

P.P.S. If you can"t get locale to work, I"d suggest a modified version of Mark"s answer:

def intWithCommas(x):
    if type(x) not in [type(0), type(0L)]:
        raise TypeError("Parameter must be an integer.")
    if x < 0:
        return "-" + intWithCommas(-x)
    result = ""
    while x >= 1000:
        x, r = divmod(x, 1000)
        result = ",%03d%s" % (r, result)
    return "%d%s" % (x, result)

Recursion is useful for the negative case, but one recursion per comma seems a bit excessive to me.

sep

How would you make a comma-separated string from a list of strings?

5 answers

mweerden By mweerden

What would be your preferred way to concatenate strings from a sequence such that between every two consecutive pairs a comma is added. That is, how do you map, for instance, ["a", "b", "c"] to "a,b,c"? (The cases ["s"] and [] should be mapped to "s" and "", respectively.)

I usually end up using something like "".join(map(lambda x: x+",",l))[:-1], but also feeling somewhat unsatisfied.

612

Answer #1

my_list = ["a", "b", "c", "d"]
my_string = ",".join(my_list)
"a,b,c,d"

This won"t work if the list contains integers


And if the list contains non-string types (such as integers, floats, bools, None) then do:

my_string = ",".join(map(str, my_list)) 

We hope this article has helped you to resolve the problem. Apart from How to avoid Python/Pandas creating an index in a saved csv?, check other _files-related topics.

Want to excel in Python? See our review of the best Python online courses 2022. If you are interested in Data Science, check also how to learn programming in R.

By the way, this material is also available in other languages:



Carlo Emmerson

Massachussetts | 2022-12-02

Thanks for explaining! I was stuck with How to avoid Python/Pandas creating an index in a saved csv? for some hours, finally got it done 🤗. I am just not quite sure it is the best method

Cornwall Krasiko

Abu Dhabi | 2022-12-02

Thanks for explaining! I was stuck with How to avoid Python/Pandas creating an index in a saved csv? for some hours, finally got it done 🤗. Checked yesterday, it works!

Davies Innsbruck

Berlin | 2022-12-02

Thanks for explaining! I was stuck with How to avoid Python/Pandas creating an index in a saved csv? for some hours, finally got it done 🤗. I just hope that will not emerge anymore

Shop

Learn programming in R: courses

$

Best Python online courses for 2022

$

Best laptop for Fortnite

$

Best laptop for Excel

$

Best laptop for Solidworks

$

Best laptop for Roblox

$

Best computer for crypto mining

$

Best laptop for Sims 4

$

Latest questions

NUMPYNUMPY

Common xlabel/ylabel for matplotlib subplots

12 answers

NUMPYNUMPY

How to specify multiple return types using type-hints

12 answers

NUMPYNUMPY

Why do I get "Pickle - EOFError: Ran out of input" reading an empty file?

12 answers

NUMPYNUMPY

Flake8: Ignore specific warning for entire file

12 answers

NUMPYNUMPY

glob exclude pattern

12 answers

NUMPYNUMPY

How to avoid HTTP error 429 (Too Many Requests) python

12 answers

NUMPYNUMPY

Python CSV error: line contains NULL byte

12 answers

NUMPYNUMPY

csv.Error: iterator should return strings, not bytes

12 answers

News


Wiki

Python | How to copy data from one Excel sheet to another

Common xlabel/ylabel for matplotlib subplots

Check if one list is a subset of another in Python

sin

How to specify multiple return types using type-hints

exp

Printing words vertically in Python

exp

Python Extract words from a given string

Cyclic redundancy check in Python

Finding mean, median, mode in Python without libraries

cos

Python add suffix / add prefix to strings in a list

Why do I get "Pickle - EOFError: Ran out of input" reading an empty file?

Python - Move item to the end of the list

Python - Print list vertically