In my program, user inputs number n, and then inputs n number of strings, which get stored in a list.
I need to code such that if a certain list index exists, then run a function.
This is made more complicated by the fact that I have nested if statements about len(my_list).
Here's a simplified version of what I have now, which isn't working:
n = input ("Define number of actors: ") count = 0 nams = [] while count < n: count = count + 1 print "Define name for actor ", count, ":" name = raw_input () nams.append(name) if nams[2]: #I am trying to say 'if nams[2] exists, do something depending on len(nams) if len(nams) > 3: do_something if len(nams) > 4 do_something_else if nams[3]: #etc. 12 Answers
Could it be more useful for you to use the length of the list len(n) to inform your decision rather than checking n[i] for each possible length?
I need to code such that if a certain list index exists, then run a function.
This is the perfect use for a try block:
ar=[1,2,3] try: t=ar[5] except IndexError: print('sorry, no 5') # Note: this only is a valid test in this context # with absolute (ie, positive) index # a relative index is only showing you that a value can be returned # from that relative index from the end of the list... However, by definition, all items in a Python list between 0 and len(the_list)-1 exist (i.e., there is no need for a try block if you know 0 <= index < len(the_list)).
You can use enumerate if you want the indexes between 0 and the last element:
names=['barney','fred','dino'] for i, name in enumerate(names): print(i + ' ' + name) if i in (3,4): # do your thing with the index 'i' or value 'name' for each item... If you are looking for some defined 'index' though, I think you are asking the wrong question. Perhaps you should consider using a mapping container (such as a dict) versus a sequence container (such as a list). You could rewrite your code like this:
def do_something(name): print('some thing 1 done with ' + name) def do_something_else(name): print('something 2 done with ' + name) def default(name): print('nothing done with ' + name) something_to_do={ 3: do_something, 4: do_something_else } n = input ("Define number of actors: ") count = 0 names = [] for count in range(n): print("Define name for actor {}:".format(count+1)) name = raw_input () names.append(name) for name in names: try: something_to_do[len(name)](name) except KeyError: default(name) Runs like this:
Define number of actors: 3 Define name for actor 1: bob Define name for actor 2: tony Define name for actor 3: alice some thing 1 done with bob something 2 done with tony nothing done with alice You can also use .get method rather than try/except for a shorter version:
>>> something_to_do.get(3, default)('bob') some thing 1 done with bob >>> something_to_do.get(22, default)('alice') nothing done with alice len(nams) should be equal to n in your code. All indexes 0 <= i < n "exist".
It can be done simply using the following code:
if index < len(my_list): print(index, 'exists in the list') else: print(index, "doesn't exist in the list") Using the length of the list would be the fastest solution to check if an index exists:
def index_exists(ls, i): return (0 <= i < len(ls)) or (-len(ls) <= i < 0) This also tests for negative indices, and most sequence types (Like ranges and strs) that have a length.
If you need to access the item at that index afterwards anyways, it is easier to ask forgiveness than permission, and it is also faster and more Pythonic. Use try: except:.
try: item = ls[i] # Do something with item except IndexError: # Do something without the item This would be as opposed to:
if index_exists(ls, i): item = ls[i] # Do something with item else: # Do something without the item I need to code such that if a certain list index exists, then run a function.
You already know how to test for this and in fact are already performing such tests in your code.
The valid indices for a list of length n are 0 through n-1 inclusive.
Thus, a list has an index i if and only if the length of the list is at least i + 1.
If you want to iterate the inserted actors data:
for i in range(n): if len(nams[i]) > 3: do_something if len(nams[i]) > 4: do_something_else ok, so I think it's actually possible (for the sake of argument):
>>> your_list = [5,6,7] >>> 2 in zip(*enumerate(your_list))[0] True >>> 3 in zip(*enumerate(your_list))[0] False You can try something like this
list = ["a", "b", "C", "d", "e", "f", "r"] for i in range(0, len(list), 2): print list[i] if len(list) % 2 == 1 and i == len(list)-1: break print list[i+1]; Oneliner:
do_X() if len(your_list) > your_index else do_something_else() Full example:
In [10]: def do_X(): ...: print(1) ...: In [11]: def do_something_else(): ...: print(2) ...: In [12]: your_index = 2 In [13]: your_list = [1,2,3] In [14]: do_X() if len(your_list) > your_index else do_something_else() 1 Just for info. Imho, try ... except IndexError is better solution.
Here's a simple, if computationally inefficient way that I felt like solving this problem today:
Just create a list of available indices in my_list with:
indices = [index for index, _val in enumerate(my_list)] Then you can test before each block of code:
if 1 in indices: "do something" if 2 in indices: "do something more" but anyone reading this should really just take the correct answer from: @user6039980
Do not let any space in front of your brackets.
Example:
n = input () ^ Tip: You should add comments over and/or under your code. Not behind your code.
Have a nice day.
1
