﻿ Python: Replace all but the last five characters of a given string into "*" and returns the new masked string - w3resource

# Python: Replace all but the last five characters of a given string into "*" and returns the new masked string

## Python Basic - 1: Exercise-90 with Solution

Write a Python program to replace all but the last five characters of a given string with "*" and returns the new string.

Sample Solution:

Python Code:

``````def new_string(str1):
return '*'*(len(str1)-5) + str1[-5:]
text = "kdi39323swe"
print("Original String: ",text)
print("new string: ",new_string(text))
text = "12345abcdef"
print("\nOriginal String: ",text)
print("new string: ",new_string(text))
text = "12345"
print("\nOriginal String: ",text)
print("new string: ",new_string(text))
``````

Sample Output:

```Original String:  kdi39323swe
new string:  ******23swe

Original String:  12345abcdef
new string:  ******bcdef

Original String:  12345
new string:  12345
```

Pictorial Presentation:

Flowchart:

Python Code Editor:

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## Python: Tips of the Day

How to make a flat list out of list of lists?

Given a list of lists l

```flat_list = [item for sublist in l for item in sublist]
```

which means:

```flat_list = []
for sublist in l:
for item in sublist:
flat_list.append(item)
```

is faster than the shortcuts posted so far. (l is the list to flatten.) Here is the corresponding function:

flatten = lambda l: [item for sublist in l for item in sublist]

As evidence, you can use the timeit module in the standard library:

```\$ python -mtimeit -s'l=[[1,2,3],[4,5,6], [7], [8,9]]*99' '[item for sublist in l for item in sublist]'
10000 loops, best of 3: 143 usec per loop
\$ python -mtimeit -s'l=[[1,2,3],[4,5,6], [7], [8,9]]*99' 'sum(l, [])'
1000 loops, best of 3: 969 usec per loop
\$ python -mtimeit -s'l=[[1,2,3],[4,5,6], [7], [8,9]]*99' 'reduce(lambda x,y: x+y,l)'
1000 loops, best of 3: 1.1 msec per loop
```

Explanation: the shortcuts based on + (including the implied use in sum) are, of necessity, O(L**2) when there are L sublists -- as the intermediate result list keeps getting longer, at each step a new intermediate result list object gets allocated, and all the items in the previous intermediate result must be copied over (as well as a few new ones added at the end). So, for simplicity and without actual loss of generality, say you have L sublists of I items each: the first I items are copied back and forth L-1 times, the second I items L-2 times, and so on; total number of copies is I times the sum of x for x from 1 to L excluded, i.e., I * (L**2)/2.

The list comprehension just generates one list, once, and copies each item over (from its original place of residence to the result list) also exactly once.

Ref: https://bit.ly/3dKsNTR