def my_abs(value):
"""Returns absolute value without using abs function"""
if value < 5 :
print(value * 1)
else:
print(value * -1)
print(my_abs(3.5))
that's my code so far but the quiz prints, for example -11.255 and 200.01 and wants the opposite for example it wants 11.255 back and -200.01
What does 5 have to do with absolute value?
Following your logic:
def my_abs(value):
"""Returns absolute value without using abs function"""
if value <= 0:
return value * -1
return value * 1
print(my_abs(-3.5))
>> 3.5
print(my_abs(3.5))
>> 3.5
Other, shorter solutions also exist and can be seen in the other answers.
The solutions so far don't take into account signed zeros. In all of them, an input of either 0.0 or -0.0 will result in -0.0.
Here is a simple and (as far as I see) correct solution:
def my_abs(value):
return (value**2)**(0.5)
A simple solution for rational numbers would be
def my_abs(value):
if value<0:
return -value
return value
Why do you want to check if value < 5?
Anyways, to replicate the abs function:
def my_abs(value):
return value if value >=0 else -1 * value
A fun one.
if (number < 0) returns a boolean but when you do math with it it's a 1 or 0. Below written as (a<0)
When the number is less than 0, we can get its positive value by adding 2 times the negative of itself. or just subtract itself twice written as (-a-a)
so when our condition fails it returns 0 which when multiplied by anything is zero so we don't add anything to our original number.
if our condition passes, we'll get a 2 times the positive so we can add to our original.
here's the code, the number you're trying to get absolute value is a in this case
a = a + (a<0) * (-a-a)
This runs slower than the built in abs() call.
I thought it was faster but it was buggy in my code when I timed it.
the fastest seem to be the if (a<0) then a = -a
num = float(input("Enter any number: "))
if num < 0 :
print("Here's the absolute value: ", num*-1)
elif num == 0 :
print("Here's the absolute value: 0")
elif num > 0 :
print("Here's the absolute value: ", num)
Related
I had the task to code the following:
Take a list of integers and returns the value of these numbers added up, but only if they are odd.
Example input: [1,5,3,2]
Output: 9
I did the code below and it worked perfectly.
numbers = [1,5,3,2]
print(numbers)
add_up_the_odds = []
for number in numbers:
if number % 2 == 1:
add_up_the_odds.append(number)
print(add_up_the_odds)
print(sum(add_up_the_odds))
Then I tried to re-code it using function definition / return:
def add_up_the_odds(numbers):
odds = []
for number in range(1,len(numbers)):
if number % 2 == 1:
odds.append(number)
return odds
numbers = [1,5,3,2]
print (sum(odds))
But I couldn’t make it working, anybody can help with that?
Note: I'm going to assume Python 3.x
It looks like you're defining your function, but never calling it.
When the interpreter finishes going through your function definition, the function is now there for you to use - but it never actually executes until you tell it to.
Between the last two lines in your code, you need to call add_up_the_odds() on your numbers array, and assign the result to the odds variable.
i.e. odds = add_up_the_odds(numbers)
def listc(favn):
num = 0
while num < favn :
num += 1
return num
list = []
i = int(raw_input("Input your favourite number : > "))
for num in range(0,i):
list.append(listc(i))
print list
The elements of the list are just same. Little iterations in code are sometime printing [None] in list also.
I want to generate a list with content as 1 to i.
There are two issues with your code.
First the while loop does not run 'favn' no. of times because the return statement is within while loop.It just runs single time, and everytime it returns 1.
Also, you should change
for num in range(0,i):
list.append(listc(i))
to
for num in range(0,i):
list.append(listc(num))
You will get the output you wanted.
If you want to generate a list from 1 to i, you can simply do list = range(1, i + 1).
I want to avoid importing different modules as that is mostly what I have found while looking online. I am stuck with this bit of code and I don't really know how to fix it or improve on it. Here's what I've got so far.
def avg(lst):
'''lst is a list that contains lists of numbers; the
function prints, one per line, the average of each list'''
for i[0:-1] in lst:
return (sum(i[0:-1]))//len(i)
Again, I'm quite new and this for loops jargon is quite confusing to me, so if someone could help me get it so the output of, say, a list of grades would be different lines containing the averages. So if for lst I inserted grades = [[95,92,86,87], [66,54], [89,72,100], [33,0,0]], it would have 4 lines that all had the averages of those sublists. I also am to assume in the function that the sublists could have any amount of grades, but I can assume that the lists have non-zero values.
Edit1: # jramirez, could you explain what that is doing differently than mine possible? I don't doubt that it is better or that it will work but I still don't really understand how to recreate this myself... regardless, thank you.
I think this is what you want:
def grade_average(grades):
for grade in grades:
avg = 0
for num in grade:
avg += num
avg = avg / len(grade)
print ("Average for " + str(grade) + " is = " + str(avg))
if __name__ == '__main__':
grades = [[95,92,86,87],[66,54],[89,72,100],[33,0,0]]
grade_average(grades)
Result:
Average for [95, 92, 86, 87] is = 90.0
Average for [66, 54] is = 60.0
Average for [89, 72, 100] is = 87.0
Average for [33, 0, 0] is = 11.0
Problems with your code: the extraneous indexing of i; the use of // to truncate he averate (use round if you want to round it); and the use of return in the loop, so it would stop after the first average. Your docstring says 'print' but you return instead. This is actually a good thing. Functions should not print the result they calculate, as that make the answer inaccessible to further calculation. Here is how I would write this, as a generator function.
def averages(gradelists):
'''Yield average for each gradelist.'''
for glist in gradelists:
yield sum(glist) /len(glist)
print(list(averages(
[[95,92,86,87], [66,54], [89,72,100], [33,0,0]])))
[90.0, 60.0, 87.0, 11.0]
To return a list, change the body of the function to (beginner version)
ret = []
for glist in gradelists:
ret.append(sum(glist) /len(glist))
return ret
or (more advanced, using list comprehension)
return [sum(glist) /len(glist) for glist in gradelists]
However, I really recommend learning about iterators, generators, and generator functions (defined with yield).
I managed to do it, some other way.
but I have a question, I had this code before
def jumphunt(start, mylist, count = 0):
if count < len(mylist):
place = mylist[start]
print(place)
if place == 0:
return True
elif start >= len(mylist) or start < 0:
return False
move_left = (start - place)
move_right = (start + place)
return jumphunt(move_right, mylist, count+1) or jumphunt(move_left, mylist, count+1)
else:
return False
but for some reason it's not trying both ways
to get to the last item on the list.
for example: [1,2,2,3,4,5,3,2,1,7,0] and ,start=mylist[0]
it supposed to jump like this (from 1-2-4-1-left to 2-left to 5-right to 0)
but it keeps trying to go right and then index is out of range etc.
I thought that if u use return of or this or that, it will try both until it reaches True, why won't it work here?
Thanks!
Include the value you want to keep as a default parameter for the method, like this:
def my_func(int, list, i=0):
a = (i + int)
if int == 0:
return True
elif a > len(list):
i -= int
else:
i += int
int = list[i]
my_func(int, list, i)
Bear in mind that it may not even always be possible to arrive at the end of the list doing the jumping pattern you describe, and even if it is possible, this method may choose the wrong branch.
A better algorithm would look like this:
def branching_search(list, start):
marks = [0]*len(list)
pos = start
while list[pos]!=0:
marks[pos]++
if marks[pos] % 2 == 0 and pos + list[pos] < len(list):
pos += list[pos]
elif marks[pos] % 2 == 1 and pos - list[pos] >= 0:
pos -= list[pos]
else:
return False
if all(item == 0 or item > 1 for item in list)
return False
return True
This way, if it comes to an item that it has already visited, it will decide to go the opposite direction that it went last time. Also, if it comes to an item that it can't leave without going out-of-bounds, or if there is not way to get to the end, it will give up and return.
EDIT: I realized there are a number of flaws in this algorithm! Although it is better than the first approach, it is not guaranteed to work, although the reasons are somewhat complicated.
Just imagine this array (the unimportant elements are left blank):
1, 2, , 5, , , , , 5, 0
The first two elements would get only one mark (thus the loop checking condition would not work), but it would still get stuck looping between the two fives.
Here is a method that will always work:
def flood_search(list):
marks = [[]]*len(list)
marks[0] = [0]
still_moving = True
while still_moving:
still_moving = False
for pos in range(0,len(list)):
if marks[pos]:
if pos + list[pos] < len(list) and not marks[pos + list[pos]]:
marks[pos + list[pos]] = marks[pos] + [list[pos]];
pos += list[pos]
still_moving = True
if pos - list[pos] >= 0 and not marks[pos - list[pos]]:
marks[pos - list[pos]] = marks[pos] + [-list[pos]];
pos -= list[pos]
still_moving = True
return marks[-1]
This works by taking every possible branch at the same time.
You can also use the method to get the actual route taken to get to the end. It can still be used as a condition, since it returns an empty list if no path is found (a falsy value), or a list containing the path if a path is found (a truthy value).
However, you can always just use list[-1] to get the last item.
I want to use a variable like this :
myVariable = 0.5
myQuery.filter(myTable.column1 == myVariable*myTable.column2)
I have then no results when I apply all() to myQuery.
If I replace the variable with its value, it is OK.
queryBidOffer = session.query(BidOffer.id, BidOffer.price, BidOffer.qmin, BidOffer.qmax)
queryBidOffer = queryBidOffer.join(Equipment).filter(BidOffer.equipment==Equipment.id, Equipment.equipment_type.in_(['L','P','W','M']))
queryBidOffer_day = queryBidOffer.filter(BidOffer.day == day)
queryBidOffer_hour = queryBidOffer_day.filter(BidOffer.start_hour == timeSlice)
queryBidOffer_hour = queryBidOffer_hour.join(EquipmentDayHour, BidOffer.equipment == EquipmentDayHour.equipment)
queryBidOffer_hour = queryBidOffer_hour.filter(EquipmentDayHour.day == day)
queryBidOffer_hour = queryBidOffer_hour.filter(EquipmentDayHour.hour == timeSlice)
factor1 = 1.00 - 0.07
queryBidOffer_hour = queryBidOffer_hour.filter(BidOffer.equipment == EquipmentDayHour.equipment, factor1*func.abs(EquipmentDayHour.ref_prog) == 930)
The problem is with the two last lines (factor1).
In the last line, if I replace factor1 by its value, it is OK.
IEEE floating point numbers as used by Python, and many databases do not always work as your school math. The reason why 0.93 worked was that that was the bitwisely exact value that you had originally stored in database. But slight error appears in the subtraction.
See this for example:
>>> print "%0.64f" % (1.0 - 0.07)
0.9299999999999999378275106209912337362766265869140625000000000000
>>> print "%0.64f" % (0.93)
0.9300000000000000488498130835068877786397933959960937500000000000
Read more here How should I do floating point comparison and then for very in depth essay you could read What Every Computer Scientist Should Know About Floating-Point Arithmetic...
One solution could be to use decimal.Decimal in Python and Numeric in SQL.
I've just tested out something similar to the code you are describing, but get results whether I use a variable or a literal:
myVariable = 0.5
myQuery.filter(myTable.column1 == myVariable*myTable.column2).all()
vs.
myQuery.filter(myTable.column1 == 0.5*myTable.column2).all()
Can you post up the literal SQL that is being genarated? i.e. the results of
print myQuery
for both of those?