My question is to write a function which returns the longest string and ignores any non-strings, and if there are no strings in the input list, then it should return None.
my answer:
def longest_string(x):
for i in max(x, key=len):
if not type(i)==str:
continue
if
return max
longest_string(['cat', 'dog', 'horse'])
I'm a beginner so I have no idea where to start. Apologies if this is quite simple.
This is how i would do it:
def longest_string(x):
Strings = [i for i in x if isinstance(i, str)]
return(max(Strings, key=len)) if Strings else None
Based on your code:
def longest_string(x):
l = 0
r = None
for s in x:
if isinstance(s, str) and len(s) > l:
l = len(s)
r = s
return r
print(longest_string([None, 'cat', 1, 'dog', 'horse']))
# horse
def longest_string(items):
try:
return max([x for x in items if isinstance(x, str)], key=len)
except ValueError:
return None
def longest_string(items):
strings = (s for s in items if isinstance(s, str))
longest = max(strings, key=len) if strings else None
return longest
print(longest_string(['cat', 'dog', 'horse']))
Your syntax is wrong (second-to-last line: if with no condition) and you are returning max which you did not define manually. In actuality, max is a built-in Python function which you called a few lines above.
In addition, you are not looping through all strings, you are looping through the longest string. Your code should instead be
def longest_string(l):
strings = [item for item in l if type(item) == str]
if len(strings):
return max(strings, key=len)
return None
You're on a good way, you could iterate the list and check each item is the longest:
def longest_string(x)
# handle case of 0 strings
if len(x) == 0:
return None
current_longest = ""
# Iterate the strings
for i in x:
# Handle nonestring
if type(i) != str:
continue
# if the current string is longer than the longest, replace the string.
if len(i) > len(current_longest):
current_longest = i
# This condition handles multiple elements where none are strings and should return None.
if len(current_longest) > 0:
return current_longest
else:
return None
Since you are a beginner, I recommend you to start using python's built-in methods to sort and manage lists. Is the best when it comes to logic and leaves less room for bugs.
def longest_string(x):
x = filter(lambda obj: isinstance(obj, str), x)
longest = max(list(x), key=lambda obj: len(obj), default=None)
return longest
Nonetheless, you were in a good way. Just avoid using python´s keywords for variable names (such as max, type, list, etc.)
EDIT: I see a lot of answers using one-liner conditionals, list comprehension, etc. I think those are fantastic solutions, but for the level of programming the OP is at, my answer attempts to document each step of the process and be as readable as possible.
First of all, I would highly suggest defining the type of the x argument in your function.
For example; since I see you are passing a list, you can define the type like so:
def longest_string(x: list):
....
This not only makes it more readable for potential collaborators but helps enormously when creating docstrings and/or combined with using an IDE that shows type hints when writing functions.
Next, I highly suggest you break down your "specs" into some pseudocode, which is enormously helpful for taking things one step at a time:
returns the longest string
ignores any non-strings
if there are no strings in the input list, then it should return None.
So to elaborate on those "specifications" further, we can write:
Return the longest string from a list.
Ignore any element from the input arg x that is not of type str
if no string is present in the list, return None
From here we can proceed to writing the function.
def longest_string(x: list):
# Immediately verify the input is the expected type. if not, return None (or raise Exception)
if type(x) != list:
return None # input should always be a list
# create an empty list to add all strings to
str_list = []
# Loop through list
for element in x:
# check type. if not string, continue
if type(element) != str:
pass
# at this point in our loop the element has passed our type check, and is a string.
# add the element to our str_list
str_list.append(element)
# we should now have a list of strings
# however we should handle an edge case where a list is passed to the function that contains no strings at all, which would mean we now have an empty str_list. let's check that
if not str_list: # an empty list evaluates to False. if not str_list is basically saying "if str_list is empty"
return None
# if the program has not hit one of the return statements yet, we should now have a list of strings (or at least 1 string). you can check with a simple print statement (eg. print(str_list), print(len(str_list)) )
# now we can check for the longest string
# we can use the max() function for this operation
longest_string = max(str_list, key=len)
# return the longest string!
return longest_string
I have a list, L. L may contain other lists inside it. These other lists may also contain lists. At some point, the indexing will return an int value. How do I make a function that returns the first encountered int value?
L = [[[1,2,3],[4,5]],[6,7]]
As you can see, L can have lists of lists until some arbitrary endpoint, of which I am unaware. I tried first doing hardcoded if loops such as:
if isinstance(L[0],int): return L[0]
elif isinstance(L[0][0],int): return L[0][0]
... #and so on
But this feels unPythonic and wrong.
Is there a built-in function or other methodology for solving this problem?
As suggested by Barmar in the comments, you could write a recursive function (the problem itself is a classical use case for recursion). However, you could equally use a while loop.
Recursive version:
def first(L):
if isinstance(L, list):
return first(L[0])
else:
return L
Version with while-loop:
def first(L):
while isinstance(L, list):
L = L[0]
return L
Instead of using isinstance, you can also use a try-block instead:
def first(L):
try:
while True:
L = L[0]
except:
return L
A last variant would use iteration instead of subscripts like so:
def first(L):
try:
while True:
L, *_ = L
except:
return L
Comparing the running times of these variants, I get:
Recursive function: 0.462
While and isinstance: 0.355
While and try: 0.319
While and iter: 0.567
So, using a while-loop with a try-block is the fastest solution. And it has the nice added benefit that it also just works out of the box with tuples instead of lists.
You can use a simple loop:
my_elem = L[0]
while not isinstance(my_elem, int):
my_elem = my_elem[0]
This is the nice part of a dynamically typed language: you can always put an int in to the same variable you were putting lists before
Sorry, it's a bit hard to explain my question in a title, but basically, I have a list of positions, and each position can be passed through a function to get a number that gives you data about the position. What I want to do is return the position with the lowest data value in the list, but I can't seem to find a way to do this.
A bit of pseudo code should help:
def posfunc(self,pos):
x,y = pos
return x**2-y
def minpos(self)
returns position with the least x**2-y value
Python is pretty cool :D:
min(positions, key=posfunc)
From built-in documentation:
>>> help(min)
min(...)
min(iterable[, key=func]) -> value
min(a, b, c, ...[, key=func]) -> value
With a single iterable argument, return its smallest item.
With two or more arguments, return the smallest argument.
And lambda's are worth mentioning here:
min(positions, key=lambda x: x[0]**2 - x[1])
Is roughly the same, but more readable I think, if you aren't using posfunc elsewhere.
you can basically use the min() function
pos = [(234, 4365), (234, 22346), (2342, 674)]
def posfunc(pos):
x,y = pos
return x**2-y
min(pos, key=posfunc)
I'm having problems with a homework question.
"Write a function, to_str(a), that takes an array, a, converts each of
its elements to a string (using str(a[i])) and appends all these
strings together."
This is what I have
def to_str(a):
for i in a: a.append([i])
return str(a[i])
I have no idea how to use str(a[i]), I was wondering if someone can point me to the right direction
From the docs:
str(object) -> string
Return a nice string representation of the object. If the argument is
a string, the return value is the same object.
So str(a[i]) will return a string representation of a[i], i.e. convert a[i] to a string.
You will then need to concatenates the strings for all values of i.
As for your code, I have the following comments:
i is an element of a, not an index, as you might be thinking;
you are appending elements of a to a (endlessly, I'm afraid);
a[i] can cause an exception, because, like I said, i is an element, not an index;
you need to return a concatenation of strings, not a string from one element.
Also, if using str(a[i]) is not strictly mandatory, I'd suggest to skip it as unpythonic. You don't need indexes at all for this. Examples:
''.join(str(element) for element in a)
or
''.join(map(str, a))
will return what you need. In both cases str is applied to all elements of a.
The simplest-to-understand ("beginner") way without using indexes will be
s = ''
for element in a:
s += str(element)
return s
It's a bit less efficient, though it does effectively the same thing.
Converting each element into a string is easiest to use list comprehension:
[ str(i) for i in a ]
# equivalent to
[ str(a[i]) for i in range(len(a)) ]
# equivalent to
map(str, a) # most concise, use if you want to feel incredibly clever...
So you can write the function:
def to_str2(a):
''.join([str(i) for i in a]) # concatenates the list as a list of strings
.
Your code nearly does this:
def to_str(a):
new_a = [] # rather than use the same a!
for i in a:
new_a.append(str(i)) #convert i to string before appending
return new_a
The code meeting all the task criteria is rather something like:
def to_str(a):
return reduce(lambda x, y: x + str(y), a, '')
Which does things exactly in the mentioned way: first converts them to strings, then adds to the string made from already processed elements (which at the beginning is just emty string).
EDIT: The clearer (and supported by Python 3) way is to use explicit looping through elements. It does exactly the same, clarifying at the same time how reduce() works:
def to_str(a):
result = '' # third argument of reduce()
for item in a:
result += str(item) # does what reduce() lambda argument was doing
return result
the simplest way is:
[str(i) for i in a]
if you want a function:
def to_str(a):
return [str(i) for i in a]
I'm calling a bunch of methods that return a list. The list may be empty. If the list is non-empty, I want to return the first item; otherwise, I want to return None. This code works:
def main():
my_list = get_list()
if len(my_list) > 0:
return my_list[0]
return None
but it seems to me that there should be a simple one-line idiom for doing this. Is there?
Python 2.6+
next(iter(your_list), None)
If your_list can be None:
next(iter(your_list or []), None)
Python 2.4
def get_first(iterable, default=None):
if iterable:
for item in iterable:
return item
return default
Example:
x = get_first(get_first_list())
if x:
...
y = get_first(get_second_list())
if y:
...
Another option is to inline the above function:
for x in get_first_list() or []:
# process x
break # process at most one item
for y in get_second_list() or []:
# process y
break
To avoid break you could write:
for x in yield_first(get_first_list()):
x # process x
for y in yield_first(get_second_list()):
y # process y
Where:
def yield_first(iterable):
for item in iterable or []:
yield item
return
The best way is this:
a = get_list()
return a[0] if a else None
You could also do it in one line, but it's much harder for the programmer to read:
return (get_list()[:1] or [None])[0]
(get_list() or [None])[0]
That should work.
BTW I didn't use the variable list, because that overwrites the builtin list() function.
The most python idiomatic way is to use the next() on a iterator since list is iterable. just like what #J.F.Sebastian put in the comment on Dec 13, 2011.
next(iter(the_list), None) This returns None if the_list is empty. see next() Python 2.6+
or if you know for sure the_list is not empty:
iter(the_list).next() see iterator.next() Python 2.2+
If you find yourself trying to pluck the first thing (or None) from a list comprehension you can switch to a generator to do it like:
next((x for x in blah if cond), None)
Pro: works if blah isn't indexable Con: it's unfamiliar syntax. It's useful while hacking around and filtering stuff in ipython though.
The OP's solution is nearly there, there are just a few things to make it more Pythonic.
For one, there's no need to get the length of the list. Empty lists in Python evaluate to False in an if check. Just simply say
if list:
Additionally, it's a very Bad Idea to assign to variables that overlap with reserved words. "list" is a reserved word in Python.
So let's change that to
some_list = get_list()
if some_list:
A really important point that a lot of solutions here miss is that all Python functions/methods return None by default. Try the following below.
def does_nothing():
pass
foo = does_nothing()
print foo
Unless you need to return None to terminate a function early, it's unnecessary to explicitly return None. Quite succinctly, just return the first entry, should it exist.
some_list = get_list()
if some_list:
return list[0]
And finally, perhaps this was implied, but just to be explicit (because explicit is better than implicit), you should not have your function get the list from another function; just pass it in as a parameter. So, the final result would be
def get_first_item(some_list):
if some_list:
return list[0]
my_list = get_list()
first_item = get_first_item(my_list)
As I said, the OP was nearly there, and just a few touches give it the Python flavor you're looking for.
Python idiom to return first item or None?
The most Pythonic approach is what the most upvoted answer demonstrated, and it was the first thing to come to my mind when I read the question. Here's how to use it, first if the possibly empty list is passed into a function:
def get_first(l):
return l[0] if l else None
And if the list is returned from a get_list function:
l = get_list()
return l[0] if l else None
New in Python 3.8, Assignment Expressions
Assignment expressions use the in-place assignment operator (informally called the walrus operator), :=, new in Python 3.8, allows us to do the check and assignment in-place, allowing the one-liner:
return l[0] if (l := get_list()) else None
As a long-time Python user, this feels like we're trying to do too much on one line - I feel it would be better style to do the presumptively equally performant:
if l := get_list():
return l[0]
return None
In support of this formulation is Tim Peter's essay in the PEP proposing this change to the language. He didn't address the first formulation, but based on the other formulations he did like, I don't think he would mind.
Other ways demonstrated to do this here, with explanations
for
When I began trying to think of clever ways to do this, this is the second thing I thought of:
for item in get_list():
return item
This presumes the function ends here, implicitly returning None if get_list returns an empty list. The below explicit code is exactly equivalent:
for item in get_list():
return item
return None
if some_list
The following was also proposed (I corrected the incorrect variable name) which also uses the implicit None. This would be preferable to the above, as it uses the logical check instead of an iteration that may not happen. This should be easier to understand immediately what is happening. But if we're writing for readability and maintainability, we should also add the explicit return None at the end:
some_list = get_list()
if some_list:
return some_list[0]
slice or [None] and select zeroth index
This one is also in the most up-voted answer:
return (get_list()[:1] or [None])[0]
The slice is unnecessary, and creates an extra one-item list in memory. The following should be more performant. To explain, or returns the second element if the first is False in a boolean context, so if get_list returns an empty list, the expression contained in the parentheses will return a list with 'None', which will then be accessed by the 0 index:
return (get_list() or [None])[0]
The next one uses the fact that and returns the second item if the first is True in a boolean context, and since it references my_list twice, it is no better than the ternary expression (and technically not a one-liner):
my_list = get_list()
return (my_list and my_list[0]) or None
next
Then we have the following clever use of the builtin next and iter
return next(iter(get_list()), None)
To explain, iter returns an iterator with a .next method. (.__next__ in Python 3.) Then the builtin next calls that .next method, and if the iterator is exhausted, returns the default we give, None.
redundant ternary expression (a if b else c) and circling back
The below was proposed, but the inverse would be preferable, as logic is usually better understood in the positive instead of the negative. Since get_list is called twice, unless the result is memoized in some way, this would perform poorly:
return None if not get_list() else get_list()[0]
The better inverse:
return get_list()[0] if get_list() else None
Even better, use a local variable so that get_list is only called one time, and you have the recommended Pythonic solution first discussed:
l = get_list()
return l[0] if l else None
Regarding idioms, there is an itertools recipe called nth.
From itertools recipes:
def nth(iterable, n, default=None):
"Returns the nth item or a default value"
return next(islice(iterable, n, None), default)
If you want one-liners, consider installing a library that implements this recipe for you, e.g. more_itertools:
import more_itertools as mit
mit.nth([3, 2, 1], 0)
# 3
mit.nth([], 0) # default is `None`
# None
Another tool is available that only returns the first item, called more_itertools.first.
mit.first([3, 2, 1])
# 3
mit.first([], default=None)
# None
These itertools scale generically for any iterable, not only for lists.
for item in get_list():
return item
Frankly speaking, I do not think there is a better idiom: your is clear and terse - no need for anything "better". Maybe, but this is really a matter of taste, you could change if len(list) > 0: with if list: - an empty list will always evaluate to False.
On a related note, Python is not Perl (no pun intended!), you do not have to get the coolest code possible.
Actually, the worst code I have seen in Python, was also very cool :-) and completely unmaintainable.
By the way, most of the solution I have seen here do not take into consideration when list[0] evaluates to False (e.g. empty string, or zero) - in this case, they all return None and not the correct element.
my_list[0] if len(my_list) else None
Not sure how pythonic this is but until there is a first function in the library I include this in the source:
first = lambda l, default=None: next(iter(l or []), default)
It's just one line (conforms to black) and avoids dependencies.
Out of curiosity, I ran timings on two of the solutions. The solution which uses a return statement to prematurely end a for loop is slightly more costly on my machine with Python 2.5.1, I suspect this has to do with setting up the iterable.
import random
import timeit
def index_first_item(some_list):
if some_list:
return some_list[0]
def return_first_item(some_list):
for item in some_list:
return item
empty_lists = []
for i in range(10000):
empty_lists.append([])
assert empty_lists[0] is not empty_lists[1]
full_lists = []
for i in range(10000):
full_lists.append(list([random.random() for i in range(10)]))
mixed_lists = empty_lists[:50000] + full_lists[:50000]
random.shuffle(mixed_lists)
if __name__ == '__main__':
ENV = 'import firstitem'
test_data = ('empty_lists', 'full_lists', 'mixed_lists')
funcs = ('index_first_item', 'return_first_item')
for data in test_data:
print "%s:" % data
for func in funcs:
t = timeit.Timer('firstitem.%s(firstitem.%s)' % (
func, data), ENV)
times = t.repeat()
avg_time = sum(times) / len(times)
print " %s:" % func
for time in times:
print " %f seconds" % time
print " %f seconds avg." % avg_time
These are the timings I got:
empty_lists:
index_first_item:
0.748353 seconds
0.741086 seconds
0.741191 seconds
0.743543 seconds avg.
return_first_item:
0.785511 seconds
0.822178 seconds
0.782846 seconds
0.796845 seconds avg.
full_lists:
index_first_item:
0.762618 seconds
0.788040 seconds
0.786849 seconds
0.779169 seconds avg.
return_first_item:
0.802735 seconds
0.878706 seconds
0.808781 seconds
0.830074 seconds avg.
mixed_lists:
index_first_item:
0.791129 seconds
0.743526 seconds
0.744441 seconds
0.759699 seconds avg.
return_first_item:
0.784801 seconds
0.785146 seconds
0.840193 seconds
0.803380 seconds avg.
try:
return a[0]
except IndexError:
return None
def head(iterable):
try:
return iter(iterable).next()
except StopIteration:
return None
print head(xrange(42, 1000) # 42
print head([]) # None
BTW: I'd rework your general program flow into something like this:
lists = [
["first", "list"],
["second", "list"],
["third", "list"]
]
def do_something(element):
if not element:
return
else:
# do something
pass
for li in lists:
do_something(head(li))
(Avoiding repetition whenever possible)
Borrowing more_itertools.first_true code yields something decently readable:
def first_true(iterable, default=None, pred=None):
return next(filter(pred, iterable), default)
def get_first_non_default(items_list, default=None):
return first_true(items_list, default, pred=lambda x: x!=default)
Following code covers several scenarios by using lambda:
l1 = [1,2,3]
l2 = []
l3 = None
first_elem = lambda x: x[0] if x else None
print(first_elem(l1))
print(first_elem(l2))
print(first_elem(l3))
Using the and-or trick:
a = get_list()
return a and a[0] or None
Probably not the fastest solution, but nobody mentioned this option:
dict(enumerate(get_list())).get(0)
if get_list() can return None you can use:
dict(enumerate(get_list() or [])).get(0)
Advantages:
-one line
-you just call get_list() once
-easy to understand
My use case was only to set the value of a local variable.
Personally I found the try and except style cleaner to read
items = [10, 20]
try: first_item = items[0]
except IndexError: first_item = None
print first_item
than slicing a list.
items = [10, 20]
first_item = (items[:1] or [None, ])[0]
print first_item
How about this:
(my_list and my_list[0]) or None
Note: This should work fine for lists of objects but it might return incorrect answer in case of number or string list per the comments below.
You could use Extract Method. In other words extract that code into a method which you'd then call.
I wouldn't try to compress it much more, the one liners seem harder to read than the verbose version. And if you use Extract Method, it's a one liner ;)
Several people have suggested doing something like this:
list = get_list()
return list and list[0] or None
That works in many cases, but it will only work if list[0] is not equal to 0, False, or an empty string. If list[0] is 0, False, or an empty string, the method will incorrectly return None.
I've created this bug in my own code one too many times !
isn't the idiomatic python equivalent to C-style ternary operators
cond and true_expr or false_expr
ie.
list = get_list()
return list and list[0] or None
if mylist != []:
print(mylist[0])
else:
print(None)