How do I append the index of a string to a list using extend() in python?

Question

I’m trying to look through a long string to find instances of a substring, then I want to create a list that has the index of each substring found and the substring found. But instead of the index in a readable form, I’m getting a reference to the object, such as [<built-in method index of str object at 0x000001687B01E930>, 'b']. I’d rather have [123, 'b'].

Here’s the code I’ve tried:

test_string = "abcdefg"
look_for = ["b","f"]
result = []

for each in test_string:
    if each in look_for:
        result.extend([each.index, each])
        
print(result)

I know I could do this with a list comprehension, but I plan to add a bunch of other code to this for later and am only asking about the index issue here.

I’ve tried str(each.index) and print(str(result))

But that doesn’t help. What am I missing?

Asked By: OutThere

||

Source

Answer 1

You can use enumerate :

test_string = "abcdefg"
look_for = ["b", "f"]
result = []

for index, each in enumerate(test_string):
    if each in look_for:
        result.extend([index, each])

print(result)

Answered By: NLavie

Answer 2

Here are 2 ways you can achieve it, either using .index() or enumerate()

test_string = "abcdefg"
look_for = ["b","f"]
result = []

for i, each in enumerate(test_string):
    if each in look_for:
        result.append((i, each))
        
print(result)

result =  []

for i in look_for:
    try:
        result.append((i, test_string.index(i)))
    except:
        pass
print(result)

Answered By: darth baba

Answer 3

This is not an answer about using .extend().

Your specification says find all substrings, but only uses a test for a single character. The following code will handle longer substrings, and do it with a single pass using regular expressions. This is a significant performance speed up as the number of substrings increases. Regular expressions are our friends!

import re
def find_all_substrings(target, substrings):
    # regex to scan for all substrings in one mass
    pattern = "|".join(map(re.escape, substrings))
    # Scan once for all strings
    matches = re.finditer(pattern, target)
    # Get start position and matched string
    results = [(match.start(),match.group()) for match in matches]
    return results

print(find_all_substrings("I have an apple, a banana, and a cherry in my bag. Also, another apple.", ['apple', 'banana', 'cherry']))
test_string = "abcdefg"
look_for = ["b", "f"]
print(find_all_substrings(test_string, look_for))

giving the following output:

[(10, 'apple'), (19, 'banana'), (33, 'cherry'), (65, 'apple')]
[(1, 'b'), (5, 'f')]

If you are determined to use a call to .extend(), possibly to allow modifying the results on the fly, you could use:

import re
def find_all_substrings_generator(target, substrings):
    # REGEX to scan for all substrings in one mass
    pattern = "|".join(map(re.escape, substrings))
    matches = re.finditer(pattern, target)
    # Get start position and matched string
    for match in matches:
        yield (match.start(), match.group())

results = []
test_string = "abcdefg"
look_for = ["b", "f"]
for i, result in enumerate(find_all_substrings_generator(test_string, look_for)):
    print(f"Result {i}: {result}")
    results.extend([result])
print(results)

With the output:

Result 0: (1, 'b')
Result 1: (5, 'f')
[(1, 'b'), (5, 'f')]

Answered By: Sean True

How do I append the index of a string to a list using extend() in python?

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