Alternating sums of a list [duplicate] - python

This question already has answers here:
Writing a function that alternates plus and minus signs between list indices
(7 answers)
Closed 2 years ago.
9.Write a program that accepts 9 integers from the user and stores them in a list. Next, compute the alternating sum of all of the elements in the list. For example, if the user enters
1 4 9 16 9 7 4 9 11
then it computes
1 – 4 + 9 – 16 + 9 – 7 + 4 – 9 + 11 = –2
myList = []
value = None
count = 0
while count != 9:
value = int(input("Please enter a number: "))
myList.append(value)
count = count + 1
if count == 9:
break
print(myList)
def newList(mylist):
return myList[0] - myList[1] + myList[2] - myList[3] + myList[4] - myList[5] + myList[6] - myList[7] + myList[8]
x = newList(myList)
print(x)
My code returns the correct answer, but I need it to print out the actual alternating sums as in the example. I have been stuck on this for a while. I am having a mental block on this and havent been able to find anything similar to this online.
I appreciate any help or tips.
Also, this is python 3.
Thank you.

a=[1, 4, 9, 16, 9, 7, 4, 9, 11]
start1=0
start2=1
sum1=0
first_list=[a[i] for i in range(start1,len(a),2)]
second_list=[a[i] for i in range(start2,len(a),2)]
string=''
for i,j in zip(first_list,second_list):
string+=str(i)+'-'+str(j)+'+'
string.rstrip('+')
print('{}={}'.format(string,str(sum(first_list)-sum(second_list))))
Output
1-4+9-16+9-7+4-9+=-2

Try doing this:
positives = myList[::2]
negatives = myList[1::2]
result = sum(positives) - sum(negatives)
print ("%s = %d" % (" + ".join(["%d - %d" % (p, n) for p, n in zip(positives, negatives)]), result))
I'll explain what I'm doing here. The first two lines are taking slices of your list. I take every other number in myList starting from 0 for positives and starting from 1 for negatives. From there, finding the result of the alternating sum is just a matter of taking the sum of positives and subtracting the sum of negatives from it.
The final line is somewhat busy. Here I zip positives and negatives together which produces a list of 2-tuples where of the form (positive, negative) and then I use string formatting to produce the p - n form. From there I use join to join these together with the plus sign, which produces p0 - n0 + p1 - n1 + p2 - n2.... Finally, I use string formatting again to get it in the form of p0 - n0 + p1 - n1 + p2 - n2 ... = result.

You can do as you did but place it in a print statement
print(myList[0] + " - " + myList[1] + " + " + myList[2] + " - " + myList[3] + " + " + myList[4] + " - " + myList[5] + " + " + myList[6] + " - " + myList[7] + " + " + myList[8] + " = " + x)
Its not perfectly clean, but it follows your logic, so your teacher won't know you got your solution from someone else.

Something along the lines of the following would work:
def sumList(theList):
value = 0
count = 0
steps = ""
for i in theList:
if count % 2 == 0:
value += i
steps += " + " + str(i)
else:
value -= i
steps += " - " + str(i)
count += 1
print(steps[3:])
return value
print(sumList(myList))
It alternates between + and - by keeping track of the place in the list and using the modulus operator. Then it calculates the value and appends to a string to show the steps which were taken.

You can also do something like below once your 9 or more numbers list is ready
st = ''
sum = 0
for i, v in enumerate(myList):
if i == 0:
st += str(v)
sum += v
elif i % 2 == 0:
st += "+" + str(v)
sum += v
else:
st += "-" + str(v)
sum -= v
print("%s=%d" % (st, sum))
It prints : 1-4+9-16+9-7+4-9+11=-2

Related

I no longer have an alignment issue, however, when I print out my X's and O's there will be a 0 next to them. Does anyone know a different solution?

This is what I have so far, if you have any ideas please let me know. It would mean a lot to me.
a_list = list(range(1, squared_input + 1))
turn = 0
Symbol_1 = "X"
Symbol_2 = "O"
while turn <= 9:
X = 1
while X < squared_input + 1 :
print(str(a_list[X - 1]).zfill(2), end= "")
if X%board_size == 0 :
print("")
print(("--+" * (board_size - 1)), end="")
print("--")
else:
print("|", end="")
X = X + 1
turn = turn + 1
Symbol_1, Symbol_2 = Symbol_2, Symbol_1
print("You are user " + Symbol_1 + ".")
user_input = input("Please pick a slot on the game board (using numbers 1 - " + str(squared_input) + "): ")
a_list[int(user_input) - 1] = Symbol_1
The zeros come from your call to zfill which explicitly pads a string with 0 to a requested size. You call zfill(2) with a string that contains a single character. So the function pads that to length two by adding a 0.
To pad with blanks you can for example use the format() function or just something like
'%2d' % a_list[X-1]
which will pad each number to length 2 from the left with blanks, or
'%-2d' % a_list[X-1]
which will pad each number to length 2 from the right with blanks.

List Index Out of Range? for Scores

I'm trying to loop through a list of teams that I attained from Patriots.com scores by using Python and BeautifulSoup. I use scrape and a loop to get the scores, teams, and Weeks for every year between 2001 and 2018. After 2001, the I get a list index out of range error
I tried changing multiple things like the integer I use or if the len() of the list needs int()
m = 0
while m < int(len(the_teams)):
m = m+1
q = the_scores[m].split('-')
val = []
for y in q:
int_val = int(y)
val.append(int_val)
if val[0] > val[1]:
print("The Patriots won against the " + the_teams[m] + " during "
+ the_game[m] + " with a score of " + the_scores[m] + "!")
else:
print("The Patriots lost against the " + the_teams[m] + " during
" +the_game[m] + " with a score of " + the_scores[m] + ".")
I expect to output every game and their score in the 2 formats in the print statements at the end
You’re incrementing your index before you’re doing your analysis, after checking if it’s a valid index.
If you move m = m+ 1 - or the more pythonic m += 1 - to the bottom of the while loop, rather than the top, it should work as expected.

How to draw this bow tie pattern using Python 2.7?

I need to draw the following pattern using Python While Loops.
I have spent quite a lot of time and came up with this code which prints it perfectly but this code is so much long and I feel like it is not one of those good codes.
If anybody here can help me out shrinking this code or suggesting a better way to output?
here is the code:
#Question 10, Alternate Approach
temp = 1
pattern = ""
innerSpace = 7
starCount = 1
while temp <= 5:
st = 1
while st <= starCount:
pattern = pattern + "*"
if st != starCount:
pattern = pattern + " "
st = st + 1
sp = 0
if temp == 5:
innerSpace = 1
while sp < innerSpace:
pattern = pattern + " "
sp = sp + 1
st = 1
while st <= starCount:
if temp == 5:
st = st + 1
pattern = pattern + "*"
if st != starCount:
pattern = pattern + " "
st = st + 1
temp = temp + 1
innerSpace = innerSpace - 2
pattern = pattern + "\n"
if temp % 2 == 0:
pattern = pattern + " "
else:
starCount = starCount + 1
starCount = 2
innerSpace = 1
while temp > 5 and temp <= 9:
st = 1
while st <= starCount:
pattern = pattern + "*"
if st != starCount:
pattern = pattern + " "
st = st + 1
sp = 0
while sp < innerSpace:
pattern = pattern + " "
sp = sp + 1
st = 1
while st <= starCount:
pattern = pattern + "*"
if st != starCount:
pattern = pattern + " "
st = st + 1
temp = temp + 1
innerSpace = innerSpace + 2
pattern = pattern + "\n"
if temp % 2 == 0:
starCount = starCount - 1
pattern = pattern + " "
print pattern
Since this looks like an assignment, I'll give you a hint how I would do it.
Take advantage of the symmetry of the bow. It is symmetrical about the horizontal and vertical axis. Therefore, you really only need to solve 1 corner, then copy/mirror the results to get the rest.
This code gives one way of looking at the problem, which is just shifting a initial string (the middle of the bow) to get the desired shape:
m = '*'
size = 4
n = 5 # must be odd
pad = ' ' * n
middle = (m + pad) * size
half = int(n / 2) + 1
print middle
print middle[half*1:]
print middle[half*2:]
print middle[half*3:]
print middle[half*4:]
print middle[half*5:]
print middle[half*6:]
Which yields this:
* * * *
* * *
* * *
* *
* *
*
*
Good luck!
I would use list comprehensions and strings and would exploit the symmetry of the figure.
Not a complete solution, but could be a part of a loop body
In [2]: a = '*' + ' '*8
In [3]: a
Out[3]: '* '
In [24]: result = ''
In [25]: result += a
In [26]: result
Out[26]: '* '
In [27]: result += a[-1::-1]
In [28]: result
Out[28]: '* *'
In [29]: result += '\n'
In [30]: a = ' '+'*' + ' '*7
In [31]: a
Out[31]: ' * '
In [32]: result += a
In [33]: result += a[-1::-1]
In [34]: result += '\n'
In [36]: print result
* *
* *
IMHO you use while loop much as if they where for loops.
I don't think that's what your teacher wants.
The idea behind while is to run until a certain condition is met, not
necessarily when the number of iterations exceed a certain limit.
The condition does not need to be included in the while statement, you can check it later and use the break command to escape the loop
Try for example this:
start = '*'
while True:
print start
if start[0] == '*':
start = ' ' + start
else:
start = '*' + start
if (start == '* * *'):
break
output is just a part of your assignment, think you should be able to work it out to the final, expected result!
Hopefully by this time HW is done. Since I solved this using dynamic programming, I thought I would list solution here.
Observations:
While looking at pattern its observed that bottom half is palindrome of top half. Hence we need to calculate only the top half.
Next we see that for every row count,we have pattern like,
row 1 = 1 , n
row 2 = 2 , n -1
row 3 = 1,3, n-2, n
row 4 = 2, 4 , n-3, n-1
.. and so on.
With iteration index as row count and n as input value we can dynamically calculate remaining values very efficiently.
Source-Code
def get_list(bound, alist):
tmp_list = []
for i in xrange(1,bound + 1):
tmp_list.append(star if i in alist else dot)
return tmp_list
star = "*"
dot = " "
n = 20 #How large of BowTie do you want?
m = (n * 2) - 1
#get top half list
th = []
for idx,k in enumerate(xrange(1,n+1)): #run through 1 - n
row = idx + 1
tmplst = []
if row % 2 != 0:
tmplst.append(i for i in xrange(1,row + 1) if i % 2 != 0)
tmplst.append(i for i in xrange(m, m-row, -1) if i % 2 != 0)
else:
tmplst.append(i for i in xrange(1,row + 1) if i % 2 == 0)
tmplst.append(i for i in xrange(m, m-row, -1) if i % 2 == 0)
#append each row value to top half list.
th.append(sorted(set([j for i in tmplst for j in i])))
#create palindrome of top half which is our bottom half
th = th + th[len(th) -2::-1]
#create list of * and blanks
final = [get_list(m, i) for i in th]
#Print BowTie
for i in final:
print ' '.join(i)
Using a stars and spacing and counting variable
counting=1
star_amount=1
space_amount=6
loop_var=7
while loop_var>0:
loop_var-=1
if space_amount==0:
counting*=-1
stars=" * "*star_amount
spaces=" "*space_amount
print(stars+spaces+stars)
star_amount+=counting
space_amount-= counting*2

Python Gtin 8 Code not totaling

Hi I have been experimenting for some time to try and total 7 variables at once. I am trying to calculate the 8th number for GTIN 8 codes. I have tried many things and so far I am using float. I Don't know what it does but people say use it. I need to times the 1,3,5,7 number by 3 and 2,4,6 number by 1. Then find the total of all of them added together. I have looked everywhere and I cant find anything. Anything will help. Thanks Ben
code = input ("enter 7 digit code? ")
sum1 = 3 * (code[0] + ',')
sum2 = code[1] + ','
sum3 = 3 * (code[2] + ',')
sum4 = code[3] + ','
sum5 = 3 * (code[4] + ',')
sum6 = code[5] + ','
sum7 = 3 * (code[6] + ',')
checksum_value = sum1 + sum2 + sum3+ sum4 + sum5+ sum6 + sum7
b = str(checksum_value)
print(b)
Quick solution:
x = "1234567"
checksum_value = sum(int(v) * 3 if i in (0,2,4,6) else int(v) for (i, v) in enumerate(x[:7]))
# (1*3) + 2 + (3*3) + 4 + (5*3) + 6 + (7*3)
# ==
# 3 + 2 + 9 + 4 + 15 + 6 + 21
# ==
# sum(int(v) * 3 if i in (0,2,4,6) else int(v) for (i, v) in enumerate(x[:7]))
Explanation:
# Sum the contained items
sum(
# multiply by three if the index is 0,2,4 or 6
int(v) * 3 if i in (0,2,4,6) else int(v)
# grab our index `i` and value `v` from `enumerate()`
for (i, v) in
# Provide a list of (index, value) from the iterable
enumerate(
# use the first 7 elements
x[:7]
)
)
`enter code here`code = input ("enter 7 digit code? ")
sum1 = 3 * (code[0] + ',')
sum2 = code[1] + ','
sum3 = 3 * (code[2] + ',')
sum4 = code[3] + ','
sum5 = 3 * (code[4] + ',')
sum6 = code[5] + ','
sum7 = 3 * (code[6] + ',')
checksum_value = sum1 + sum2 + sum3+ sum4 + sum5+ sum6 + sum7
b = str(checksum_value)
print(b)
GS1 codes come in different lengths, ranging from GTIN-8 (8 digits) to SSCC (2 digit application ID + 18 digits). Here's a simple, general Python formula that works for any length GS1 identifier:
cd = lambda x: -sum(int(v) * [3,1][i%2] for i, v in enumerate(str(x)[::-1])) % 10
Explanation:
Convert input to string, so input can be numeric or string - just a convenience factor.
Reverse the string - simple way to align the 3x/1x pattern with variable-length input.
The weighting factor is selected based on odd and even input character position by calculating i mod 2. The last character in the input string (i=0 after the string has been reversed) gets 3x.
Calculate the negative weighted sum mod 10. Equivalent to the (10 - (sum mod 10)) mod 10 approach you'd get if you follow the GS1 manual calculation outline exactly, but that's ugly.
Test Cases
## GTIN-8
>>> cd(1234567)
0
>>> cd(9505000)
3
## GTIN-12
>>> cd(71941050001)
6
>>> cd('05042833241')
2
## GTIN-13
>>> cd(900223631103)
6
>>> cd(501234567890)
0
## GTIN-14
>>> cd(1038447886180)
4
>>> cd(1001234512345)
7
## SSCC (20 digits incl. application identifier)
>>> cd('0000718908562723189')
6
>>> cd('0037612345000001009')
1

Function doesn't return all results from 'for' loop [duplicate]

This question already has answers here:
How can I use `return` to get back multiple values from a loop? Can I put them in a list?
(2 answers)
Closed 6 months ago.
I've made a simple function to print out a times table chart depending on the number you decide to run with. The problem I'm having due to my basic understanding of the language is why it only returns the first loop and nothing else.
def timestables(number):
for a in range(1, number+1):
b = a*a
c = a
return (str(c) + " * " + str(c) + " = " + str(b))
print(timestables(5))
I get the answer ..
1 * 1 = 1
I've tried to rectify this issue by using print instead of return but this ultimately results with a None appearing as well.
def timestables(number):
for a in range(1, number+1):
b = a*a
c = a
print (str(c) + " * " + str(c) + " = " + str(b))
print(timestables(5))
I get the answer ..
1 * 1 = 1
2 * 2 = 4
3 * 3 = 9
4 * 4 = 16
5 * 5 = 25
None
How can I return all given results from the for loop to avoid a None error?
yield them.
def timestables(number):
for a in range(1, number+1):
yield '%s + %s = %s' % (a, a, a*a )
for x in timestables(5):
print x
This turns your function into a generator function, and you need to iterate over the results, i.e. the result is not a list, but an iterable.
If you need a list, the simplest is to explicitly create one:
res = list(timestables(5))
But again, if you don't, you don't.
IMHO, this is the most pythonic way.
You're returning inside the for loop - and functions stop execution immediately once they hit a return statement.
To work around this, you can use a list to store those values, and then return that list.
def timestables(number):
lst = []
for a in range(1, number+1):
b = a*a
c = a
lst.append(str(c) + " * " + str(c) + " = " + str(b))
return lst
As a side note, you should use string formatting to build the string, like so.
lst.append('{a} * {a} = {b}'.format(a=a, b=a*a))
Now we can get rid of all those intermediate variables (b and c), and we can use a list comprehension instead.
def timestables(number):
return ['{a} * {a} = {b}'.format(a=a, b=a*a) for a in range(1, number+1)]
If you don't want the function to return a list, but a multi-line string, you can use str.join:
def timestables(number):
return '\n'.join('{a} * {a} = {b}'.format(a=a, b=a*a) for a in range(1, number+1))
Now we can test the function:
>>> print(timestables(5))
1 * 1 = 1
2 * 2 = 4
3 * 3 = 9
4 * 4 = 16
5 * 5 = 25
You can return an array:
def timestables(number):
out = []
for a in range(1, number+1):
b = a*a
c = a
out.append( str(c) + " * " + str(c) + " = " + str(b))
return out
print(timestables(5))

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