How do you calculate a negative number in python?

I am a student in a concepts of programming class. The lab is run by a TA and today in lab he gave us a real simple little program to build. It was one where it would multiply by addition. Anyway, he had us use absolute to avoid breaking the prog with negatives. I whipped it up real quick and then argued with him for 10 minutes that it was bad math. It was, 4 * -5 does not equal 20, it equals -20. He said that he really dosen't care about that and that it would be too hard to make the prog handle the negatives anyway. So my question is how do I go about this.

here is the prog I turned in:

#get user input of numbers as variables numa, numb = input("please give 2 numbers to multiply seperated with a comma:") #standing variables total = 0 count = 0 #output the total while (count< abs(numb)): total = total + numa count = count + 1 #testing statements if (numa, numb <= 0): print abs(total) else: print total

I want to do it without absolutes, but every time I input negative numbers I get a big fat goosegg. I know there is some simple way to do it, I just can't find it.

asked Mar 16, 2010 at 2:37

3

Perhaps you would accomplish this with something to the effect of

text = raw_input("please give 2 numbers to multiply separated with a comma:") split_text = text.split(',') a = int(split_text[0]) b = int(split_text[1]) # The last three lines could be written: a, b = map(int, text.split(',')) # but you may find the code I used a bit easier to understand for now. if b > 0: num_times = b else: num_times = -b total = 0 # While loops with counters basically should not be used, so I replaced the loop # with a for loop. Using a while loop at all is rare. for i in xrange(num_times): total += a # We do this a times, giving us total == a * abs(b) if b < 0: # If b is negative, adjust the total to reflect this. total = -total print total

or maybe

a * b

answered Mar 16, 2010 at 2:52

Mike GrahamMike Graham

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Too hard? Your TA is... well, the phrase would probably get me banned. Anyways, check to see if numb is negative. If it is then multiply numa by -1 and do numb = abs(numb). Then do the loop.

answered Mar 16, 2010 at 2:50

2

The abs() in the while condition is needed, since, well, it controls the number of iterations (how would you define a negative number of iterations?). You can correct it by inverting the sign of the result if numb is negative.

So this is the modified version of your code. Note I replaced the while loop with a cleaner for loop.

#get user input of numbers as variables numa, numb = input("please give 2 numbers to multiply seperated with a comma:") #standing variables total = 0 #output the total for count in range(abs(numb)): total += numa if numb < 0: total = -total print total

answered Mar 16, 2010 at 3:03

slackerslacker

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Try this on your TA:

# Simulate multiplying two N-bit two's-complement numbers # into a 2N-bit accumulator # Use shift-add so that it's O(base_2_log(N)) not O(N) for numa, numb in ((3, 5), (-3, 5), (3, -5), (-3, -5), (-127, -127)): print numa, numb, accum = 0 negate = False if numa < 0: negate = True numa = -numa while numa: if numa & 1: accum += numb numa >>= 1 numb <<= 1 if negate: accum = -accum print accum

output:

3 5 15 -3 5 -15 3 -5 -15 -3 -5 15 -127 -127 16129

answered Mar 16, 2010 at 4:13

John MachinJohn Machin

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How about something like that? (Uses no abs() nor mulitiplication)
Notes:

• the abs() function is only used for the optimization trick. This snippet can either be removed or recoded.
• the logic is less efficient since we're testing the sign of a and b with each iteration (price to pay to avoid both abs() and multiplication operator)

def multiply_by_addition(a, b): """ School exercise: multiplies integers a and b, by successive additions. """ if abs(a) > abs(b): a, b = b, a # optimize by reducing number of iterations total = 0 while a != 0: if a > 0: a -= 1 total += b else: a += 1 total -= b return total multiply_by_addition(2,3) 6 multiply_by_addition(4,3) 12 multiply_by_addition(-4,3) -12 multiply_by_addition(4,-3) -12 multiply_by_addition(-4,-3) 12

answered Mar 16, 2010 at 3:13

mjvmjv

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Thanks everyone, you all helped me learn a lot. This is what I came up with using some of your suggestions

#this is apparently a better way of getting multiple inputs at the same time than the #way I was doing it text = raw_input("please give 2 numbers to multiply separated with a comma:") split_text = text.split(',') numa = int(split_text[0]) numb = int(split_text[1]) #standing variables total = 0 if numb > 0: repeat = numb else: repeat = -numb #for loops work better than while loops and are cheaper #output the total for count in range(repeat): total += numa #check to make sure the output is accurate if numb < 0: total = -total print total

Thanks for all the help everyone.

answered Mar 16, 2010 at 23:42

dman762000dman762000

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Try doing this:

num1 = int(input("Enter your first number: ")) num2 = int(input("Enter your second number: ")) ans = num1*num2 if num1 > 0 or num2 > 0: print(ans) elif num1 > 0 and num2 < 0 or num1 < 0 and num1 > 0: print("-"+ans) elif num1 < 0 and num2 < 0: print("Your product is "+ans) else: print("Invalid entry")

answered Dec 2, 2021 at 2:46

import time print ('Two Digit Multiplication Calculator') print ('===================================') print () print ('Give me two numbers.') x = int ( input (':')) y = int ( input (':')) z = 0 print () while x > 0: print (':',z) x = x - 1 z = y + z time.sleep (.2) if x == 0: print ('Final answer: ',z) while x < 0: print (':',-(z)) x = x + 1 z = y + z time.sleep (.2) if x == 0: print ('Final answer: ',-(z)) print ()

answered Nov 10, 2014 at 20:34

KitiyoKitiyo

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