﻿ Python: Remove all instances of a given value from a given array of integers and find the length of the new array - w3resource

# Python: Remove all instances of a given value from a given array of integers and find the length of the new array

## Python Basic - 1: Exercise-75 with Solution

Write a Python program to remove all instances of a given value from a given array of integers and find the length of the new array.

Sample Solution:

Python Code:

``````def remove_element(array_nums, val):
i = 0
while i < len(array_nums):
if array_nums[i] == val:
array_nums.remove(array_nums[i])

else:
i += 1

return len(array_nums)
print(remove_element([1, 2, 3, 4, 5, 6, 7, 5], 5))
print(remove_element([10,10,10,10,10], 10))
print(remove_element([10,10,10,10,10], 20))
print(remove_element([], 1))
``````

Sample Output:

```6
0
5
0
```

Pictorial Presentation:

Flowchart:

Python Code Editor:

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## Python: Tips of the Day

Iterating over dictionaries using 'for' loops:

I am a bit puzzled by the following code: d = {'x': 1, 'y': 2, 'z': 3} for key in d: print key, 'corresponds to', d[key] What I don't understand is the key portion. How does Python recognize ...

key is just a variable name.

```for key in d:
```

For Python 3.x:

```for key, value in d.items():
```

For Python 2.x:

```for key, value in d.iteritems():
```

To test for yourself, change the word key to poop.

In Python 3.x, iteritems() was replaced with simply items(), which returns a set-like view backed by the dict, like iteritems() but even better. This is also available in 2.7 as viewitems().

The operation items() will work for both 2 and 3, but in 2 it will return a list of the dictionary's (key, value) pairs, which will not reflect changes to the dict that happen after the items() call. If you want the 2.x behavior in 3.x, you can call list(d.items()).

Ref: https://bit.ly/37dm0Qo