﻿ Python: Find the first appearance of the substring 'not' and 'poor' from a given string, if 'not' follows the 'poor', replace the whole 'not'...'poor' substring with 'good'. Return the resulting string - w3resource

# Python: Find the first appearance of the substring 'not' and 'poor' from a given string, if 'not' follows the 'poor', replace the whole 'not'...'poor' substring with 'good'. Return the resulting string

## Python String: Exercise-7 with Solution

Write a Python program to find the first appearance of the substring 'not' and 'poor' from a given string, if 'not' follows the 'poor', replace the whole 'not'...'poor' substring with 'good'. Return the resulting string.

Sample Solution:-

Python Code:

``````def not_poor(str1):
snot = str1.find('not')
spoor = str1.find('poor')

if spoor > snot and snot>0 and spoor>0:
str1 = str1.replace(str1[snot:(spoor+4)], 'good')
return str1
else:
return str1
print(not_poor('The lyrics is not that poor!'))
print(not_poor('The lyrics is poor!'))
```
```

Sample Output:

```The lyrics is good!
The lyrics is poor!
```

Flowchart:

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## Python: Tips of the Day

Check if a given key already exists in a dictionary:

In is the intended way to test for the existence of a key in a dict.

```d = {"key1": 10, "key2": 23}

if "key1" in d:
print("this will execute")

if "nonexistent key" in d:
print("this will not")
```

If you wanted a default, you can always use dict.get():

```d = dict()

for i in range(100):
key = i % 10
d[key] = d.get(key, 0) + 1
```

and if you wanted to always ensure a default value for any key you can either use dict.setdefault() repeatedly or defaultdict from the collections module, like so:

```from collections import defaultdict

d = defaultdict(int)

for i in range(100):
d[i % 10] += 1
```

but in general, the in keyword is the best way to do it.

Ref: https://bit.ly/2XPMRyz