2

I'm sure this is simple and I'm looking right at it, but I cannot make it work.

Goal: Use a loop to go through the list of strings and find the search term. Return the element number of the first match.

I've tried a few options, nothing seems to work, and I have yet to find a working description of how to do it in any texts.

This is my best attempt so far: 

def get_element_number(a_list, search_term):
    for i in range(len(a_list)):
      if search_term in a_list[i]:
        return a_list.index(i)
      elif not search_term in a_list:
        return 'no match'

Error message:

Traceback (most recent call last):
File "python", line 11, in <module>
File "python", line 5, in get_element_number
ValueError: 2 is not in list

Not looking for the complete answer, just any help in where I'm going wrong or if I'm missing something would be very helpful.

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2
  • Can you provide a sample of a_list and search_term as made in the function call? Commented Jan 5, 2017 at 12:21
  • oh sorry! get_element_number(['a'', 'b', 'd'], 'd') I found a work around that it'll accept using the length of the list, but I definitely appreciate the help and real answers! Commented Jan 5, 2017 at 12:57

6 Answers 6

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3

You can replace all of this with index = a_list.index(search_term).

Note that if the list doesn't contain the search_term, it will throw an exception, so you'll need to catch that and return "not found" or something similar. Second note: it only returns the first index of the found search_term.

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1 Comment

I figured out already what to do for the exception (basically broke down the challenge into parts, solved what I could and banged my head on the desk for a bit over the rest). Thank you for the help!
3

if search_term in a_list[i]: is True even if search_term is contained in a_list[i].

So in the case of an exact match index works, but in a case of a partial match index throws an exception.

Aside: elif not search_term in a_list: is wrong. Remove it or you'll return at first non-match.

Rewrite it as:

def get_element_number(a_list, search_term):
     try:
          return a_list.index(search_term)
     except IndexError:
          'no match'

This is simpler and has the advantage of performing search only once, which is important performance-wise when you're using linear search (not taking the overhead of exception into account).

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Comments

0 
for index, s in enumerate(a_list):
  if s == term:
    return index
return -1
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Comments

0 
def get_element_number(a_list, search_term):
    for index, value in enumerate(a_list):
        if search_term in value:
            return index
    return 'not match'
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1 Comment

I was reading about enumerate, but I thought it worked a little different. Thank you!
0

Here's a simple change to your code that will resolve it:

def get_element_number(a_list, search_term):
    if search_term in a_list: #if search keyword is in the list
        return a_list.index(search_term) #then return the index of that
    else:
        return 'no match!'
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Comments

0 
def find_element(a_list, search_term):
i = 0
for e in a_list:
    if e == search_term:
        return i
    i += 1
return -1

Very nice explanation for either this one or other solution you can find in the video linked below

https://www.youtube.com/watch?time_continue=216&v=GwOFHuPMCpM

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2 Comments

"This is if you want to be consist in for loop." isn't a grammatically correct sentence -- can you clarify?
I meant "remain consistent" as he posted the issue using for loop to solve that problem.

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