python - checking odd even numbers and changing outputs on number size

Question

I have a couple of problems to solve for an assignment  and am a bit stuck  The question is to write a program that gets the user to input an odd number  check it s odd   then print an upside down pyramid of stars based on the size of the input   For example  if you enter 5  it comes up with                   My problem is therefore two-fold   1  How do I check if it s even or odd  I tried if number 2    int in the hope that it might do something  and the internet tells me to do if number 2  0  but that doesn t work   2  How do I change the asterisks in the middle of each line   Any help with either problem is greatly appreciated

User · Answer

My solution basically we have two string and with the & we get the right index:

res = ["Even", "Odd"]
print(res[x & 1])

Please note that it seems slower than other alternatives:

#!/usr/bin/env python3
import math
import random
from timeit import timeit

res = ["Even", "Odd"]

def foo(x):
    return res[x & 1]

def bar(x):
    if x & 1:
        return "Odd"
    return "Even"

la = lambda x : "Even" if not x % 2 else "Odd"

iter = 10000000

time = timeit('bar(random.randint(1, 1000))', "from __main__ import bar, random", number=iter)
print(time)
time = timeit('la(random.randint(1, 1000))', "from __main__ import la, random", number=iter)
print(time)
time = timeit('foo(random.randint(1, 1000))', "from __main__ import foo, random", number=iter)
print(time)

output:
8.05739480999182
8.170479692984372
8.892275177990086

User · Answer

The modulo 2 solutions with  2 is good  but that requires a division and a subtraction   Because computers use binary arithmetic  a much more efficient solution is     This first solution does not produce a Boolean value   is odd if zero   value  amp  1    or  is odd    value  amp  1     1    or  is even    value  amp  1     0

User · Answer

A few of the solutions here reference the time taken for various  is even  operations  specifically n   2 vs n  amp  1   without systematically checking how this varies with the size of n  which turns out to be predictive of speed   The short answer is that if you re using reasonably sized numbers  normally  lt  1e9  it doesn t make much difference   If you re using larger numbers then you probably want to be using the bitwise operator   Here s a plot to demonstrate what s going on  with Python 3 7 3  under Linux 5 1 2      Basically as you hit  arbitrary precision  longs things get progressively slower for modulus  while remaining constant for the bitwise op   Also  note the 10  -7 multiplier on this  i e  I can do  30 million  small integer  checks per second   Here s the same plot for Python 2 7 16     which shows the optimisation that s gone into newer versions of Python   I ve only got these versions of Python on my machine  but could rerun for other versions of there s interest   There are 51 ns between 1 and 1e100  evenly spaced on a log scale   for each point I do the equivalent of   timeit  n   2   f n  n    number niter     where niter is calculated to make timeit take  0 1 seconds  and this is repeated 5 times   The slightly awkward handling of n is to make sure we re not also benchmarking global variable lookup  which is slower than local variables   The mean of these values are used to draw the line  and the individual values are drawn as points

User · Answer

Modulus method is the usual method  We can also do this to check if odd or even   def f a       if  a  2  2    a          return  even      else          return  odd    Integer division by 2 followed by multiplication by two

User · Answer

if number 2  0   will tell you that it s even  So odd numbers would be the else statement there  The     is the mod sign which returns the remainder after dividing  So essentially we re saying if the number is divisible by two we can safely assume it s even  Otherwise it s odd  it s a perfect correlation    As for the asterisk placing you want to prepend the asterisks with the number of spaces correlated to the line it s on  In your example        line 0       line 1       line 2   We ll want to space accordingly  0      01    012

User · Answer

Giving you the complete answer would have no point at all since this is homework  so here are a few pointers    Even or Odd   number   2    0   definitely is a very good way to find whether your number is even    In case you do not know    this does modulo which is here the remainder of the division of number by 2  http   en wikipedia org wiki Modulo operation  Printing the pyramid   First advice  In order to print        you can do print       5   Second advice  In order to center the asterisks  you need to find out how many spaces to write before the asterisks  Then you can print a bunch of spaces and asterisks with print     1       3

User · Answer

Regarding the printout  here s how I would do it using the Format Specification Mini Language  section  Aligning the text and specifying a width    Once you have your length  say length   11   rowstring          length d     format length   length    center aligned  space-padded format string of length  lt length gt  for i in xrange length  0  -2     iterate from top to bottom with step size 2     print rowstring format        i

User · Answer

there are a lot of ways to check if an int value is odd or even  I ll show you the two main ways   number   5  def best way number       if number 2  0          print  even      else          print  odd   def binary way number       if str bin number   len bin number  -1    0           print  even      else          print  odd  best way number  binary way number    hope it helps

User · Answer

This is simple code  You can try it and grab the knowledge easily   n   int input  Enter integer           if n   2    3 8huhubuiiujji           print  digit entered is ODD       elif n   2    0 and 2  lt  n  lt  5          print  EVEN AND in between  2 5        elif n   2    0 and 6  lt  n  lt  20          print  EVEN and in between  6 20        elif n   2    0 and n  gt  20         print  Even and greater than 20     and so on

User · Answer

Simple but yet fast    gt  gt  gt  def is odd a           return bool a -   a gt  gt 1  lt  lt 1        gt  gt  gt  print is odd 13   True  gt  gt  gt  print is odd 12   False  gt  gt  gt    Or even simpler    gt  gt  gt  def is odd a         return bool a  amp  1

User · Answer

Sample Instruction Given an integer  n  performing the following conditional actions    If n is odd  print Weird If n is even and in the inclusive range of 2 to 5  print Not Weird If n is even and in the inclusive range of 6 to 20  print Weird If n is even and greater than 20  print Not Weird   import math n   int input     if n   2   1      print  Weird   elif n   2  0 and n in range 2 6       print  Not Weird   elif n   2    0 and n in range 6 21       print  Weird   elif n   2  0 and n gt 20      print  Not Weird

User · Answer

def main        n   float input  odd         while n   2    0           if n   2    1  No need for these lines as if it were true the while loop would not have been entered               break not required as the while condition will break loop         n   float input  odd           for i in range int  n 1  2            print     i     int  n-2 i       i   main      1st part ensures that it is an odd number that was entered 2nd part does the printing of triangular

User · Answer

la   lambda x    even  if not x   2 else  odd

User · Answer

1  another odd testing function Ok  the assignment was handed in 8  years ago  but here is another solution based on bit shifting operations  def isodd i       return bool i gt  gt 0 amp 1    testing gives   gt  gt  gt  isodd 2  False  gt  gt  gt  isodd 3  True  gt  gt  gt  isodd 4  False  2  Nearest Odd number alternative approach However  instead of a code that says  quot give me this precise input  an integer odd number  or otherwise I won t do anything quot  I also like robust codes that say   quot give me a number  any number  and I ll give you the nearest pyramid to that number quot   In that case this function is helpful  and gives you the nearest odd  e g  any number f such that 6 lt  f lt 8 is set to 7 and so on   def nearodd f       return int f 2  2 1  Example output  nearodd 4 9  5  nearodd 7 2  7  nearodd 8  9

User · Answer

Here s my solution   def is even n       r n 2 0     return True if r  int r  else False

User · Answer

I guess the easiest and most basic way is this  import math  number   int  input   Enter number       if number   2    0 and number    0      print   Even number   elif number    0      print   Zero is neither even  nor odd    else      print   Odd number     Just basic conditions and math  It also minds zero  which is neither even  nor odd and you give any number you want by input so it s very variable

User · Answer

1  How do I check if it s even or odd  I tried  if number 2    int  in the hope that it might do something  and the internet tells me to do  if number 2  0   but that doesn t work       def isEven number           return number   2    0

[python] python - checking odd/even numbers and changing outputs on number size

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