So if I have a list, let's say: [1,2,'hello'] and I want to add all the numbers in a list, (strings should be ignored). I want it to return the number 3 (1+2). And the function must be recursive.
This is what I have come up with:
import numbers
def isnumber(x):
return isinstance(x, numbers.Number)
def sum_first(list):
if not list:
return 0
elif isnumber(list[-1]):
return list[-1] + sum_first(list[:-1])
list=eval(input("Enter a list: "))
print(sum_first(list))
This only works if the last element in the list is a number. Let's assume it's a string, how could I change it?
You are only handling two cases:
the list is empty
the last element is a number
Your example list doesn't fit either case, so you end up returning None.
Add a third branch:
def sum_first(lst):
if not lst:
return 0
elif isnumber(lst[-1]):
return lst[-1] + sum_first(lst[:-1])
else:
return sum_first(lst[:-1])
Now you handle the case where the last element is not a number; you handle it by ignoring it; recursing with the list without that one element.
I renamed your list variable to lst to avoid masking the built-in type.
reduce can be seen as a form of recursive function. If you don't intend to write your own recursive function then may be you could do this
import operator as op
def silent_int(v):
try:
return int(v)
except ValueError:
return 0
xs = [1, 2, 'hello', '4']
print reduce(op.add, map(silent_int, xs))
Obviously the correct way of counting numbers should be using sum but with the recursion restriction you could use reduce or build your own trampoline!
Related
I would like print a result without duplicate with my multiplication
Here an example :
5*3*2=30
2*3*5=30
5*2*3=30
3*2*5=30
.....
All these element are from my list that I browse and you can see it's always =30
So I would like display only the first element (5*3*2) and not others because they are the same.
To be more accurate, here an example what I have :
list = ['5*3*2','5*2*3','2*3*5','2*5*3']
for i in list:
if eval(i) == eval(i)+1 ??? (I dont know how to say the next element)
print(eval(i))
Thanks for reading
Something like this with not in will help you.
#python 3.5.2
list = ['5*3*2','5*2*3','6*9','2*3*5','2*5*3','8*3','9*6']
elm = []
for i in list:
elm_value = eval(i)
if elm_value not in elm:
elm.append(elm_value)
print(elm)
DEMO: https://rextester.com/QKV22875
The comparison:
eval(i) == eval(i)+1
Will compare if the the number i is equal to i + 1, which will always return False. I'm sure you mean to use i as an index and simply wanted to compare if the current element is equal to the next element in the list. However, doing this doesn't really keep track of duplicates, since you have to consider everything else in the list.
Its also not a good idea to use list as a variable name, since it shadows the builtin function list. Plenty of other suitable names you can use.
One way is to use a set to keep track of what items you have seen, and only print items that you have seen for the first time:
lst = ["5*3*2","5*2*3","2*3*5","2*5*3"]
seen = set()
for exp in lst:
calc = eval(exp)
if calc not in seen:
print(calc)
seen.add(calc)
If you are always dealing with simple multiplying expressions with the * operator(no brackets), you could also use functools.reduce and operator.mul instead to multiply the numbers instead of eval here. This will first split the numbers by *, map each number string to an integer, then multiply every element with each other.
from operator import mul
from functools import reduce
lst = ["5*3*2","5*2*3","2*3*5","2*5*3"]
seen = set()
for exp in lst:
numbers = map(int, exp.split("*"))
calc = reduce(mul, numbers)
if calc not in seen:
print(calc)
seen.add(calc)
Output:
30
With the following list:
l = ['5*3*2','5*2*3','2*3*5','2*5*3', '2*2']
(Note that list is already something in python so I wouldn't recommend using that as a variable name)
I would first create a list of unique values:
unique_vals = set(map(eval, list))
set([4, 30])
Then for each unique values get the first match in l:
[next(x for x in l if eval(x) == i) for i in unique_vals]
I get:
['2*2', '5*3*2']
Is that what you want?
def lengthgood(x):
for i in x:
if len(i)<13
return i
else:
pass
def makeproperlist(x):
return x.split(',')
attendancelist=makeproperlist(input("attendee list:"))
final_list=list(filter(lengthgood,attendancelist))
for i in finallist:
print (i)
I want to write a programme of which I create a list in which only elements shorter than 14 can be a part of.
This is my code, but it keeps returning all the elements I put in, even if some of them are longer than 14?
I tried to print the len(i) and it says that it is 1 for every element in the list?
How do I solve this?
You shouldn't put a return within a loop; it'll only return the first element that matches your condition.
You also shouldn't be looping within your filter function, because you're actually looping over characters of strings, which are all length 1.
Therefore, the first character is always returning a truthy value, giving you back the initial input after filtering
You only need to check the input length, and filter functions should ideally return appropriate boolean conditions rather than truthy values (in your case return i is returning a non-empty string)
def lengthgood(x):
return len(x)<13
If you don't need to use filter(), you can write a list comprehension
final_list=[a if len(a) < 13 for a in attendancelist]
may be like that
flst = []
def croper(lst):
for i in lst:
flst.append(i) if len(i) < 13 else 0
lst = input("attendee list:").split(',')
croper(lst)
print(flst)
or shorter
def croper(lst):
return [i for i in lst if len(i) < 13]
lst = input("attendee list:").split(',')
print(croper(lst))
You want to create a list but you not define it anywhere in program simply done it with simply one function makeproperlist(x).
Try this code bro this will help you.
attendancelist=[]
while (True):
ch=input("Enter c for continue and e for exit : ")
if (ch=='c'):
def makeproperlist(x):
if x<=13:
attendancelist.append(x)
else:
print("Enter number which is <= to 13.")
makeproperlist(int(input("attendee list:")))
elif (ch=='e'):
break;
print("Your attendece list : ",attendancelist)
I am learning the recursive functions. I completed an exercise, but in a different way than proposed.
"Write a recursive function which takes a list argument and returns the sum of its integers."
L = [0, 1, 2, 3, 4] # The sum of elements will be 10
My solution is:
def list_sum(aList):
count = len(aList)
if count == 0:
return 0
count -= 1
return aList[0] + list_sum(aList[1:])
The proposed solution is:
def proposed_sum(aList):
if not aList:
return 0
return aList[0] + proposed_sum(aList[1:])
My solution is very clear in how it works.
The proposed solution is shorter, but it is not clear for me why does the function work. How does if not aList even happen? I mean, how would the rest of the code fulfill a not aList, if not aList means it checks for True/False, but how is it True/False here?
I understand that return 0 causes the recursion to stop.
As a side note, executing without if not aList throws IndexError: list index out of range.
Also, timeit-1million says my function is slower. It takes 3.32 seconds while the proposed takes 2.26. Which means I gotta understand the proposed solution.
On the call of the function, aList will have no elements. Or in other words, the only element it has is null. A list is like a string or array. When you create a variable you reserve some space in the memory for it. Lists and such have a null on the very last position which marks the end so nothing can be stored after that point. You keep cutting the first element in the list, so the only thing left is the null. When you reach it you know you're done.
If you don't use that condition the function will try to take a number that doesn't exist, so it throws that error.
You are counting the items in the list, and the proposed one check if it's empty with if not aList this is equals to len(aList) == 0, so both of you use the same logic.
But, you're doing count -= 1, this has no sense since when you use recursion, you pass the list quiting one element, so here you lose some time.
According to PEP 8, this is the proper way:
• For sequences, (strings, lists, tuples), use the fact that empty
sequences are false.
Yes: if not seq:
if seq:
No: if len(seq)
if not len(seq)
Here is my amateur thougts about why:
This implicit check will be faster than calling len, since len is a function to get the length of a collection, it works by calling an object's __len__ method. This will find up there is no item to check __len__.
So both will find up there is no item there, but one does it directly.
not aList
return True if there is no elements in aList. That if statement in the solution covers edge case and checks if input parameter is not empty list.
For understand this function, let's run it step by step :
step 0 :
L=[0,1,2,3,4]
proposed_sum([0,1,2,3,4])
L != []
return l[0] + proposed_sum([1,2,3,4])
step 1 calcul proposed_sum([1,2,3,4]):
proposed_sum([1,2,3,4])
L != []
return l[0] + sum([2,3,4])
step 2 calcul proposed_sum([2,3,4]):
proposed_sum([2,3,4])
L != []
return l[0] + sum([3,4])
step 3 calcul proposed_sum([3,4]):
proposed_sum([3,4])
L != []
return l[0] + sum([4])
step 4 calcul proposed_sum([4]):
proposed_sum([4])
L != []
return l[0] + sum([])
step 5 calcul proposed_sum([]):
proposed_sum([])
L == []
return 0
step 6 replace:
proposed_sum([0,1,2,3,4])
By
proposed_sum([]) + proposed_sum([4]) + proposed_sum([3,4]) + proposed_sum([2,3,4]) + proposed_sum([1,2,3,4])+ proposed_sum([0,1,2,3,4])
=
(0) + 4 + 3 + 2 + 1 + 0
Python considers as False multiple values:
False (of course)
0
None
empty collections (dictionaries, lists, tuples)
empty strings ('', "", '''''', """""", r'', u"", etc...)
any other object whose __nonzero__ method returns False
in your case, the list is evaluated as a boolean. If it is empty, it is considered as False, else it is considered as True. This is just a shorter way to write if len(aList) == 0:
in addition, concerning your new question in the comments, consider the last line of your function:
return aList[0] + proposed_sum(aList[1:])
This line call a new "instance" of the function but with a subset of the original list (the original list minus the first element). At each recursion, the list passed in argument looses an element and after a certain amount of recursions, the passed list is empty.
I have a function here:
def evenlengthchecker(nestedlist):
length = len(nestedlist[0])
for element in nestedlist:
if len(element) != length:
return False
This actually does work when the given nested list contains values. However, when I try something like evenlengthchecker([]), IndexErrors everywhere!
The problem is that your code starts by checking the list at position 0, which is an index error with an empty list. Here's an alternative method that won't give an error:
return (len(set(len(elt) for elt in nestedlist)) <= 1)
This just checks if there is more than one list length in the set of lengths; if you end up with the empty set, no harm done.
def evenlengthchecker(nestedlist):
a = [len(i) for i in nestedlist]
return len(set(a)) ==1
you can use all:
return all(len(x)==len(my_list[0]) for x in my_list)
Python's list type has an index(x) method. It takes a single parameter x, and returns the (integer) index of the first item in the list that has the value x.
Basically, I need to invert the index(x) method. I need to get the index of the first value in a list that does NOT have the value x. I would probably be able to even just use a function that returns the index of the first item with a value != None.
I can think of a 'for' loop implementation with an incrementing counter variable, but I feel like I'm missing something. Is there an existing method, or a one-line Python construction that can handle this?
In my program, the situation comes up when I'm handling lists returned from complex regex matches. All but one item in each list have a value of None. If I just needed the matched string, I could use a list comprehension like '[x for x in [my_list] if x is not None]', but I need the index in order to figure out which capture group in my regex actually caused the match.
Exiting at the first match is really easy: instead of computing a full list comprehension (then tossing away everything except the first item), use next over a genexp. Assuming for example that you want -1 when no item satisfies the condition of being != x,
return next((i for i, v in enumerate(L) if v != x), -1)
This is Python 2.6 syntax; if you're stuck with 2.5 or earlier, .next() is a method of the genexp (or other iterator) and doesn't accept a default value like the -1 above (so if you don't want to see a StopIteration exception you'll have to use a try/except). But then, there is a reason more releases were made after 2.5 -- continuous improvement of the language and its built-ins!-)
Using a list comprehension when you only need the first just feels slimy (to me). Use a for-loop and exit early.
>>> lst = [None, None, None, "foo", None]
>>> for i, item in enumerate(lst):
... if item: break
... else:
... print "not found"
...
>>> i
3
enumerate() returns an iterator that yields a tuple of the current index of the iterable as well as the item itself.
[i for i, x in enumerate(my_list) if x != value][0]
If you're not sure whether there's a non-matching item, use this instead:
match = [i for i, x in enumerate(my_list) if x != value]
if match:
i = match[0]
# i is your number.
You can make this even more "functional" with itertools, but you will soon reach the point where a simple for loop is better. Even the above solutions aren't as efficient as a for loop, since they construct a list of all non-matching indices before you pull the one of interest.
A silly itertools-based solution:)
import itertools as it, operator as op, functools as ft
def index_ne(item, sequence):
sequence= iter(sequence)
counter= it.count(-1) # start counting at -1
pairs= it.izip(sequence, counter) # pair them
get_1st= it.imap(op.itemgetter(0), pairs) # drop the used counter value
ne_scanner= it.ifilter(ft.partial(op.ne, item), get_1st) # get only not-equals
try:
ne_scanner.next() # this should be the first not equal
except StopIteration:
return None # or raise some exception, all items equal to item
else:
return counter.next() # should be the index of the not-equal item
if __name__ == "__main__":
import random
test_data= [0]*20
print "failure", index_ne(0, test_data)
index= random.randrange(len(test_data))
test_data[index]= 1
print "success:", index_ne(0, test_data), "should be", index
All this just to take advantage of the itertools.count counting :)