numpy.delete () in Python

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About:
numpy.delete (array, object, axis = None): returns a new array with subarrays deleted along with the mentioned axis.
Parameters :

  array:  [array_like] Input array.  object:  [int, array of ints] Sub-array to delete  axis:  Axis along which we want to delete sub-arrays. By default, it object is applied to applied to flattened array 

Return:

 An array with sub-array being deleted as per the mentioned object along a given axis. 

Code 1: removing from 1D array

# Python program illustrating
# numpy.delete ()

import numpy as geek


# Working on 1D

arr = geek.arange ( 5 )

print ( "arr:" , arr)

print ( "Shape:" , arr.shape)


# removing from 1D array

object = 2

a = geek.delete (arr, object )

print ( "deleteing arr 2 times:" , a)

print ( " Shape: " , a.shape)

object = [ 1 , 2 ]

b = geek.delete (arr, object )

print ( "deleteing arr 3 times:" , b)

print ( "Shape:" , a.shape)

Output:

 arr: [0 1 2 3 4 ] Repeating arr 2 times: [0 0 1 1 2 2 3 3 4 4] Shape: (10,) Repeating arr 3 times: [0 0 0 ..., 4 4 4] Shape: (15,) 

Code 2:

# Python program illustrating
# numpy.delete ()

import numpy as geek


# Working on 1D

arr = geek.arange ( 12 ). reshape ( 3 , 4 )

print ( "arr:" , arr)

print ( "Shape : " , arr.shape)


# remove from 2D array

a = geek.delete (arr, 1 , 0 )

"" "

[[0 1 2 3]

[4 5 6 7] - & gt; removed

[8 9 10 11]]

"" "

print ( "deleteing arr 2 times:" , a)

print ( "Shape:" , a.shape)


# remove from 2D array

a = geek.delete (arr, 1 , 1 )

"" "

[[0 1 * 2 3 ]

[4 5 * 6 7]

[8 9 * 10 11]]

^

deletion

"" "

print ( "deleteing arr 2 times:" , a)

print ( " Shape: " , a.shape)

Output:

 arr: [[0 1 2 3] [4 5 6 7] [8 9 10 11]] Shape: (3, 4) deleteing arr 2 times: [[0 1 2 3] [8 9 10 11] ] Shape: (2, 4) deleteing arr 2 times: [[0 2 3] [4 6 7] [8 10 11]] Shape: (3, 3) deleteing arr 3 times: [0 3 4 5 6 7 8 9 10 11] Shape: (3, 3) 

Code 3: deletion is performed using boolean m asks

# Python program illustrating
# numpy.delete ()

import numpy as geek

arr = geek.arange ( 5 )

print ( "Original array:" , arr)

mask = geek.ones ( len (arr), dtype = bool )


# Equivalent to np.delete (arr, [0,2,4 ], axis = 0)

mask [[ 0 , 2 ]] = False

print ( " Mask set as: " , mask)

result = arr [mask, ...]

print ( "Deletion Using a Boolean Mask:" , result)

Output:

 Original array : [0 1 2 3 4] Mask set as: [False True False True True] Deletion Using a Boolean Mask: [1 3 4] 

Links:
https://docs.scipy.org/ doc / numpy / reference / generated / numpy.delete.html

Notes:
These codes will not work for online IDs. Please run them on your systems to see how they work

This article is provided by Mohit Gupta_OMG

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numpy.delete () in Python __del__: Questions

How can I make a time delay in Python?

5 answers

I would like to know how to put a time delay in a Python script.

2973

Answer #1

import time
time.sleep(5)   # Delays for 5 seconds. You can also use a float value.

Here is another example where something is run approximately once a minute:

import time
while True:
    print("This prints once a minute.")
    time.sleep(60) # Delay for 1 minute (60 seconds).

2973

Answer #2

You can use the sleep() function in the time module. It can take a float argument for sub-second resolution.

from time import sleep
sleep(0.1) # Time in seconds

numpy.delete () in Python __del__: Questions

How to delete a file or folder in Python?

5 answers

How do I delete a file or folder in Python?

2639

Answer #1


Path objects from the Python 3.4+ pathlib module also expose these instance methods:

2639

Answer #2


Path objects from the Python 3.4+ pathlib module also expose these instance methods:

2639

Answer #3

Python syntax to delete a file

import os
os.remove("/tmp/<file_name>.txt")

Or

import os
os.unlink("/tmp/<file_name>.txt")

Or

pathlib Library for Python version >= 3.4

file_to_rem = pathlib.Path("/tmp/<file_name>.txt")
file_to_rem.unlink()

Path.unlink(missing_ok=False)

Unlink method used to remove the file or the symbolik link.

If missing_ok is false (the default), FileNotFoundError is raised if the path does not exist.
If missing_ok is true, FileNotFoundError exceptions will be ignored (same behavior as the POSIX rm -f command).
Changed in version 3.8: The missing_ok parameter was added.

Best practice

  1. First, check whether the file or folder exists or not then only delete that file. This can be achieved in two ways :
    a. os.path.isfile("/path/to/file")
    b. Use exception handling.

EXAMPLE for os.path.isfile

#!/usr/bin/python
import os
myfile="/tmp/foo.txt"

## If file exists, delete it ##
if os.path.isfile(myfile):
    os.remove(myfile)
else:    ## Show an error ##
    print("Error: %s file not found" % myfile)

Exception Handling

#!/usr/bin/python
import os

## Get input ##
myfile= raw_input("Enter file name to delete: ")

## Try to delete the file ##
try:
    os.remove(myfile)
except OSError as e:  ## if failed, report it back to the user ##
    print ("Error: %s - %s." % (e.filename, e.strerror))

RESPECTIVE OUTPUT

Enter file name to delete : demo.txt
Error: demo.txt - No such file or directory.

Enter file name to delete : rrr.txt
Error: rrr.txt - Operation not permitted.

Enter file name to delete : foo.txt

Python syntax to delete a folder

shutil.rmtree()

Example for shutil.rmtree()

#!/usr/bin/python
import os
import sys
import shutil

# Get directory name
mydir= raw_input("Enter directory name: ")

## Try to remove tree; if failed show an error using try...except on screen
try:
    shutil.rmtree(mydir)
except OSError as e:
    print ("Error: %s - %s." % (e.filename, e.strerror))

Is there a simple way to delete a list element by value?

5 answers

I want to remove a value from a list if it exists in the list (which it may not).

a = [1, 2, 3, 4]
b = a.index(6)

del a[b]
print(a)

The above case (in which it does not exist) shows the following error:

Traceback (most recent call last):
  File "D:zjm_codea.py", line 6, in <module>
    b = a.index(6)
ValueError: list.index(x): x not in list

So I have to do this:

a = [1, 2, 3, 4]

try:
    b = a.index(6)
    del a[b]
except:
    pass

print(a)

But is there not a simpler way to do this?

1055

Answer #1

To remove an element"s first occurrence in a list, simply use list.remove:

>>> a = ["a", "b", "c", "d"]
>>> a.remove("b")
>>> print(a)
["a", "c", "d"]

Mind that it does not remove all occurrences of your element. Use a list comprehension for that.

>>> a = [10, 20, 30, 40, 20, 30, 40, 20, 70, 20]
>>> a = [x for x in a if x != 20]
>>> print(a)
[10, 30, 40, 30, 40, 70]

We hope this article has helped you to resolve the problem. Apart from numpy.delete () in Python, check other __del__-related topics.

Want to excel in Python? See our review of the best Python online courses 2023. 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:



Oliver Danburry

Munchen | 2023-02-01

I was preparing for my coding interview, thanks for clarifying this - numpy.delete () in Python in Python is not the simplest one. Will get back tomorrow with feedback

Angelo Sikorski

London | 2023-02-01

I was preparing for my coding interview, thanks for clarifying this - numpy.delete () in Python in Python is not the simplest one. I am just not quite sure it is the best method

Angelo Lehnman

Warsaw | 2023-02-01

I was preparing for my coding interview, thanks for clarifying this - numpy.delete () in Python in Python is not the simplest one. Checked yesterday, it works!

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