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

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

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:

numpy.delete () in Python __delete__: Questions

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]

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