While solving problem 32 of project eular, I made this function to check pandigital number, is this a right way to check pandigital number? I think there is some problem, but I can't figure out it myself.
def pandigital(number, digit): # digit: n-digit pandigital number
l = ['{}'.format(j) for j in [i for i in range(1,digit+1)]]
g = list(str(number))
g.sort()
if g == l:
return True
Yes, there is a problem. In fact, three of them.
First, if a number contains more than one digit of the same denomination (duplicates), then g is not equal to l because l has only one copy of each digit. You should convert l and g to sets before the comparison.
Second, your function does not return anything if the number is not pandigital.
Finally, range(1,digit+1) does not include 0 (unless you want zeroless pandigital numbers).
Consider the following solution:
def pandigital(number, digit)
return {f'{i}' for i in range(digit)} == set(str(number))
Replace f'{i}' with '{}'.format(i) if you want to stick to more "classical" Python.
Related
I'm novice programmer.
I want the smallest of the input values to be output, but I don't know what's wrong.
Input example :
10
10 4 2 3 6 6 7 9 8 5
Output example :
2
n = int(input())
a = input().split()
min=a[0]
for i in range(n) :
if a[i] < min :
min = a[i]
print(min)
what is the problem? please help me
Your code should work (and it does for me).
Nevertheless, min is a reserved Python word. Taking that into consideration, I also recommend the following changes for it to be more idiomatic:
a = input().split()
min_num = a[0]
for element in a:
if element < min :
min = element
print(min)
Variables can be either number or strings. For example "2" is different from 2.
The function split returns an array of strings. You would need to convert each to a number if you want to do number comparison, like:
n = int(input())
a = input().split()
min=int(a[0])
for i in range(n) :
if int(a[i]) < min :
min = int(a[i])
print(min)
Note: you already did that for n (first line in original code), but you did not do the same when you access a.
So, min is actually a python built-in, which would be very useful in this scenario. Also, you are not making the input values in a into integers, so we can do that too:
n = int(input())
a = list(map(int, input().split()))
print(min(a))
We use map to turn all the values from the split list into integers, then turn the map object back into a list. Then, using min, we can find the smallest number very easily, without a for loop.
I think you should convert each element to integers before comparing them.
a = [int(i) for i in input().split()]
Your code should work, but it will compare strings against strings instead of integers against integers.
a= int(input())
# I input 12345
b = a
list(map(int, b))
print (list[0]*2+list[3]*1)
#can't seem to get 6 as my answer
how do I attain my answer? I can't seem to call the elements in the list. Thank you for your help.
Since you're treating the input as individual digits, you should avoid converting the input to an integer as a whole, but map the individual digits to integers as a sequence of characters:
a= input()
b = list(map(int, a))
print(b[0] * 2 + b[3] * 1)
There are several reasons why your code won't work, including your use of the map function, the fact that you do not assign the result to a variable and the use of list (which is a keyword in Python).
However, consider this code snippet which calculates your desired output:
a = int(input('Enter a number: '))
b = [int(digit) for digit in str(a)]
res = 2 * b[0] + b[3]
print(res)
Basically you have to transform your integer into a string to be able to iterate over it. Afterwards you create your list of digits out of it and can do your calculations.
Generally speaking, you should learn the basics of Python properly. A good starting point would be the official documentation (LINK).
I have two lists. I have to choose one. I have to choose the one with the smallest nth element. So I can choose the smallest element easy with min, but how do I back track it to the list itself. Have literally no idea how to solve this presumably easy problem.
a = [2,45,1,56]
b= [0,23,3,87]
Which list has the smallest element at position 2? The answer here is list a.
In case I wasnt clear, the program sould be able to solve this task for any pair of lists.
Here is a very simple snippet that does what you want, but you might want to check for the size of the arrays, in case the index is out of range.
def choose_smallest(a, b, i):
if len(a) >= i or len(b) >= i:
return 0 # do whatever you want here
if a[i] < b[i]:
return a
else:
return b
Also notice that both nth elements in your array can have the exact same value... In this example array b will be returned, but you can change that behaviour if needed.
EDIT
Added array length check
According to your example, here is a sample code you can try. You can change the code as per your requirement.
a = [2,45,1,56]
b = [0,23,3,87]
n= int(input('Enter element number: ')) # n starts from zero to length of list - 1
if a[n] > b[n]:
print('List b has smaller nth element')
elif a[n] < b[n]:
print('List a has smaller nth element')
else:
print('Both lists have equal nth element')
So we have to write a function in python to count how many digits of a non-negative integer are odd numbers.
def odd_dig(n):
ans = 0
for i in range(n):
if i in range(n) %2 == 1:
ans += 1
elif n[i]==0:
return None
I don't know python, but for a solution, why not something like the following?
while n > 0 do these two things:
add (n%2) to your count
divide n by 10
because n is defined as non-negative we should be fine with this.
You're using range(n) which creates a list from 1 to n. So when you do range(5), in your for loop, you're iterating over 1,2,3,4,5 instead of the actual digits of n.
If you want to find the number of digits you can do something like len(str(n)) which finds the length of the string form of n.
After that, make sure you're using the right operators on the right variables (string operators on strings, list operators on lists, etc...).
I'm working on my final for a class I'm taking(Python 3) im stuck at this part.
he gave us a file with numbers inside of it. we opened it and add those numbers to a list.
"Create a function called makeOdd() that returns an integer value. This function should take in any integer and reduce it down to an odd number by dividing it in half until it becomes an odd number.
o For example 10 would be cut in half to 5.
o 9 is already odd, so it would stay 9.
o But 12 would be cut in half to 6, and then cut in half again to 3.
o While 16 would be cut to 8 which gets cut to 4 which gets cut to 2 which gets cut to 1.
Apply this function to every number in the array. "
I have tried to search the internet but i have not clue where to even begin with this one. any help would be nice.
Here my whole final so far:
#imports needed to run this code.
from Final_Functions import *
#Defines empty list
myList = []
sumthing = 0
sortList = []
oddList = []
count = 0
#Starts the Final Project with my name,class, and quarter
intro()
print("***************************************************************",'\n')
#Opens the data file and reads it then places the intrager into a list we can use later.
with open('FinalData.Data', 'r') as f:
myList = [line.strip() for line in f]
print("File Read Complete",'\n')
#Finds the Sum and Adverage of this list from FinalData.Data
print("*******************sum and avg*********************************")
for oneLine in myList:
tempNum = int(oneLine)
sumthing = sumthing + tempNum
avg = sumthing /1111
print("The Sum of the List is:",sumthing)
print("The Adverage of the List is:",avg,'\n')
print("***************************************************************",'\n')
#finds and prints off the first Ten and the last ten numbers in the list
firstTen(myList)
lastTen(myList)
print("***************************************************************",'\n')
#Lest sort the list then find the first and last ten numbers in this list
sortList = myList
sortList.sort()
firstTen(sortList)
lastTen(sortList)
print("****************************************************************",'\n')
Language:Python 3
I don't want to give you the answer outright, so I'm going to talk you through the process and let you generate your own code.
You can't solve this problem in a single step. You need to divide repeatedly and check the value every time to see if it's odd.
Broadly speaking, when you need to repeat a process there are two ways to proceed; looping and recursion. (Ok, there are lots, but those are the most common)
When looping, you'd check if the current number x is odd. If not, halve it and check again. Once the loop has completed, x will be your result.
If using recursion, have a function that takes x. If it's odd, simply return x, otherwise call the function again, passing in x/2.
Either of those methods will solve your problem and both are fundamental concepts.
adding to what #Basic said, never do import * is a bad practice and is a potential source of problem later on...
looks like you are still confuse in this simple matter, you want to given a number X reduce it to a odd number by dividing it by 2, right? then ask yourself how I do this by hand? the answer is what #Basic said you first ask "X is a even number?" if the answer is No then I and done reducing this number, but if the answer is Yes then the next step dividing it by 2 and save the result in X, then repeat this process until you get to the desire result. Hint: use a while
to answer your question about
for num in myList:
if num != 0:
num = float(num)
num / 2
the problem here is that you don't save the result of the division, to do that is as simple as this
for num in myList:
if num != 0:
num = float(num)
num = num / 2