I've just started learning python and have just finished writing a program that allows a user to find the nth term of the Fibonacci sequence. I have tested it and it appears to work as intended. My question is how can I make my code more concise or more pythonic. Any advice or tips would be greatly appreciated.
values = [0,1]
n1 = 2
while True: # input and loop to limit the input range.
try:
n = int(input('Please enter position of fibonacci sequence you wish to know.'))
break
except:
print('That\'s not a valid input')
if n < 3: #calculation block
print (n)
else:
while n1 != n:
n2 = values[0] + values[1]
values[0] = values[1]
values[1] = n2
n1 +=1
print (n2)
Using functions would make your code more "pythonic" - Conciseness may not be your first priority:
def get_user_input():
while True: # input and loop to limit the input range.
try:
n = int(input('Please enter position of fibonacci sequence you wish to know.'))
return n
except:
print('That\'s not a valid input')
def get_nth_fib_number(nth):
values = [0, 1]
current_fib = 2
while current_fib <= nth:
values.append(values[-2] + values[-1])
current_fib += 1
return values[nth]
get_nth_fib_number(get_user_input())
Related
Basically, I'm trying to build a code to get the largest number from the user's inputs. This is my 1st time using a for loop and I'm pretty new to python. This is my code:
session_live = True
numbers = []
a = 0
def largest_num(arr, n):
#Create a variable to hold the max number
max = arr[0]
#Using for loop for 1st time to check for largest number
for i in range(1, n):
if arr[i] > max:
max = arr[i]
#Returning max's value using return
return max
while session_live:
print("Tell us a number")
num = int(input())
numbers.insert(a, num)
a += 1
print("Continue? (Y/N)")
confirm = input()
if confirm == "Y":
pass
elif confirm == "N":
session_live = False
#Now I'm running the function
arr = numbers
n = len(arr)
ans = largest_num(arr, n)
print("Largest number is", ans)
else:
print(":/")
session_live = False
When I try running my code this is what happens:
Tell us a number
9
Continue? (Y/N)
Y
Tell us a number
8
Continue? (Y/N)
Y
Tell us a number
10
Continue? (Y/N)
N
Largest number is 9
Any fixes?
The error in your largest_num function is that it returns in the first iteration -- hence it will only return the larger of the first two numbers.
Using the builtin max() function makes life quite a bit easier; any time you reimplement a function that already exists, you're creating work for yourself and (as you've just discovered) it's another place for bugs to creep into your program.
Here's the same program using max() instead of largest_num(), and removing a few unnecessary variables:
numbers = []
while True:
print("Tell us a number")
numbers.append(int(input()))
print("Continue? (Y/N)")
confirm = input()
if confirm == "Y":
continue
if confirm == "N":
print(f"Largest number is {max(numbers)}")
else:
print(":/")
break
So, first things first,
the use of max can be avoided, as it is a reserved keyword in python
And coming to your fix, you are comparing it with the value only once in the loop, and you are returning the number, the indentation is the key here. You will have to wait for the loop to complete its job then return the value.
There are many inbuilt methods to do the job, Here is your implementation (a bit modified)
session_live = True
numbers = []
a = 0
def largest_num(arr, n):
#Create a variable to hold the max number
max_number = arr[0]
#Using for loop for 1st time to check for largest number
for i in range(1, n):
if arr[i] > max_number:
max_number = arr[i]
# --- The indentation matters
#Returning max's value using return
return max_number
while session_live:
print("Tell us a number")
num = int(input())
numbers.insert(a, num)
a += 1
print("Continue? (Y/N)")
confirm = input()
if confirm == "Y":
pass
elif confirm == "N":
session_live = False
#Now I'm running the function
arr = numbers
n = len(arr)
ans = largest_num(arr, n)
print("Largest number is", ans)
else:
print(":/")
session_live = False
I made it without using the built-in function 'max'.
It is a way to update the 'maxNum' variable with the largest number by comparing through the for statement.
numbers = []
while True:
print("Tell us a number")
numbers.append(int(input()))
print("Continue? (Y/N)")
confirm = input()
if confirm == "Y":
continue
if confirm == "N":
maxNum = numbers[0]
for i in numbers:
if i > maxNum:
maxNum = i
print("Largest number is", maxNum)
else:
print(":/")
break
I know how to do this with a while loop and know how to use a for-loop in other languages like Java and C++. I want to use a for-loop in place of where I have written the while loop asking for the user input.
# You are required to use for-loop to solve this and round your answer to 2 decimal places. Write
# a program that takes n ∈ N (i.e., any positive integer including zero) from the user and use the
# input value to compute the sum of the following series:
n = -1
while n < 0:
n = int(input("Enter a value to compute: "))
# keep asking for user input until a whole number (0, 1, 2, 3, etc...) has been entered
k = 0
sum = 0
# To hold the sum of the fraction to be displayed
lastTerm = 0
# This variable represents the last term to be added to the fraction sum before the while loop below terminates
if n == 0:
sum = 0
elif n == 1:
sum = 1
else:
while lastTerm != 1 / n:
lastTerm = (n - k) / (k + 1)
sum = sum + (n - k) / (k + 1)
k += 1
print("{:.2f}".format(sum))
# Print the sum to two decimal places
One option is to catch the exception which is thrown when you cannot convert the input to an int, i.e.
while(True):
try:
# read input and try and covert to integer
n = int(input("Enter a value to compute: "))
# if we get here we got an int but it may be negative
if n < 0:
raise ValueError
# if we get here we have a non-negative integer so exit loop
break
# catch any error thrown by int()
except ValueError:
print("Entered value was not a postive whole number")
Alternative, slightly cleaner but I'm not 100% sure isdigit() will cover all cases
while(true):
n = input("Enter a value to compute: ")
if value.isdigit():
break
else:
print("Entered value was not a postive whole number")
How about this? It uses the for loop and sums all the values in the list.
x=[1,2,3,4] #== test list to keep the for loop going
sum_list=[]
for i in x:
j=float(input("Enter a number: "))
if not j.is_integer() or j<0:
sum_list.append(j)
x.append(1) #=== Add element in list to keep the cyclone going
else:
break
sums=sum(sum_list)
print("The Sum of all the numbers is: ",round(sums,2))
Use this to check for whole numbers -
if num < 0:
# Not a whole number
elif num >= 0:
# A whole number
for a for loop:
import itertools
for _ in itertools.repeat([]): # An infinite for loop
num = input('Enter number : ')
if num < 0:
# Not a whole number
pass # This will ask again
elif num >= 0:
# A whole number
break # break from for loop to continue the program
Easier Way -
mylist = [1]
for i in mylist : # infinite loop
num = int(input('Enter number : '))
if num < 0:
mylist.append(1)
pass # This will ask again
elif num >= 0:
# A whole number
break
I am trying to learn Python basics. I am designing a code that would print a triangle, code posted below. The expected output of the program must be a triangle from the right side not from the left which part of the code to modify to get the expected output.
The output of the program should not be like below
I want the output to be like this
Any help on this is highly appreciated. Thanks
while True:
n = int(input("enter a number between 0 and 10:"))
if 0 <= n <= 10:
break
print('try again')
rows = n
for num in range(rows+1):
for i in range(1, num+1):
if num % 2 == 0:
#print(end="" '#')
print('#',end="")
else:
#print(num)
print(num,end="")
print(" ")
We can use a list to create the elements (when it's an odd number just append the value in number times otherwise character '#' number times as the description said )of the triangle and then use a for loop to print them as below:
while True:
n = int(input("enter a number between 0 and 10:"))
if 0 <= n <= 10:
break
print('try again')
result = [idx*'#' if idx % 2 == 0 else f'{idx}'*idx for idx in range(1, n+1) ]
for i in range(len(result)):
print(' '*(n-i-1), result[i])
The result would be:
1
##
333
####
55555
Hope this can help you :)
Checking your code with the input (5) gives the output (1##333####55555 )
but the following code for the same input (5) gives the output (55555####333##1 )
while True:
n = int(input("enter a number between 0 and 10:"))
if 0 <= n <= 10:
break
print('try again')
rows = n
for num in range(rows, 0,-1):
for i in range(num+1, 1,-1):
if num % 2 == 0:
#print(end="" '#')
print('#',end="")
else:
#print(num)
print(num,end="")
print(" ")
Hope this answer your question
We have to find out the average of a list of numbers entered through keyboard
n=0
a=''
while n>=0:
a=input("Enter number: ")
n+=1
if int(a)==0:
break
print(sum(int(a.list()))/int(n))
You are not saving the numbers entered. Try :
n = []
while True:
a=input("Enter number: ")
try: #Checks if entered data is an int
a = int(a)
except:
print('Entered data not an int')
continue
if a == 0:
break
n.append(a)
print(sum(n)/len(n))
Where the list n saves the entered digits as a number
You need to have an actual list where you append the entered values:
lst = []
while True:
a = int(input("Enter number: "))
if a == 0:
break
else:
lst.append(a)
print(sum(lst) / len(lst))
This approach still has not (yet) any error management (a user enters float numbers or any nonsense or zero at the first run, etc.). You'd need to implement this as well.
a needs to be list of objects to use sum, in your case its not. That is why a.list doens't work. In your case you need to take inputs as int (Can be done like: a = int(input("Enter a number")); ) and then take the integer user inputs and append to a list (lets say its name is "ListName")(listName.append(a)), Then you can do this to calculate the average:
average = sum(listName) / len(listName);
def calc_avg():
count = 0
sum = 0
while True:
try:
new = int(input("Enter a number: "))
if new < 0:
print(f'average: {sum/count}')
return
sum += new
count += 1
print(f'sum: {sum}, numbers given: {count}')
except ValueError:
print("That was not a number")
calc_avg()
You can loop, listen to input and update both s (sum) and c (count) variables:
s, c = 0, 0
while c >= 0:
a = int(input("Enter number: "))
if a == 0:
break
else:
s += a
c += 1
avg = s/c
print(avg)
I am beginner in programming, So can you please tell me what's wrong with my code?
I want to print next palindrome number if the number entered by the user (n) is not palindrome
n = int(input("Enter any number :- "))
reverse = 0
temp = n
while (n!=0):
reverse = reverse * 10
reverse = reverse + n%10
n=n//10
if(temp==reverse):
print ("Already palindrome:: ")
if(temp != reverse):
new_temp = temp
new_reverse = 0
for i in range(new_temp,new_temp+10):
while(temp != 0):
new_reverse = new_reverse * 10
new_reverse = new_reverse + temp%10
temp = temp//10
if(new_temp==new_reverse):
print ("Next pallindrome is :- ",new_temp)
break
if(new_temp != new_reverse):
temp = new_temp+1
There are two problems with your code.
1) Your "for i in range" loop calculates the reverse of the temp variable, but you don't change the temp variable's value.
You do
new_temp = temp
for i in range(new_temp,new_temp+10):
[SNIP]
if(new_temp != new_reverse):
temp = new_temp+1 #this value never changes.
So you're making 10 iterations with one and the same value.
2) Ten iterations might not be enough to find a palindrome. Keep going until you find a palindrome.
Working code:
def reverse(num):
reverse= 0
while num:
reverse= reverse*10 + num%10
num= num//10
return reverse
num= int(input("Enter any number :- "))
if num==reverse(num):
print ("Already palindrome.")
else:
while True:
num+= 1
if num==reverse(num):
print ("Next palindrome is : %s"%num)
break
To check if a number is a palindrome, you don't need to convert it to a number. In fact, its a lot simpler if you just check the string equivalent of your number.
>>> i = '212'
>>> i == i[::-1]
True
>>> i = '210'
>>> i == i[::-1]
False
Use this to your advantage, and create a function:
def is_palindrome(foo):
return str(foo) == str(foo)[::-1]
Next, to find the next palindrome, simply increment the number till your palindrome check is true.
Combine all that, and you have:
def is_palindrome(n):
return str(n) == str(n)[::-1]
n = raw_input('Enter a number: ')
if is_palindrome(n):
print('Congratulations! {0} is a palindrome.'.format(n))
else:
n1 = n
while not is_palindrome(n1):
n1 = int(n1)+1
print('You entered {0}, but the next palindrome is {1}'.format(n, n1))
Here is how it works:
$ python t.py
Enter a number: 123
You entered 123, but the next palindrome is 131
$ python t.py
Enter a number: 121
Congratulations! 121 is a palindrome.
If it helps, I believe it's possible to solve this problem with n/2 iterations where n is the length of the input number. Here's my solution in Python:
def next_palin_number(number):
number+=1
# Convert the number to a list of its digits.
number = list(str(number))
# Initialize two indices for comparing symmetric digits.
i = 0
j = len(number) - 1
while i < j:
# If the digits are different:
if number[i] != number[j]:
# If the lower-power digit is greater than the higher-power digit:
if int(number[j]) > int(number[i]):
if number[j-1]!='9':
number[j - 1] = str(int(number[j - 1]) + 1)
number[j] = number[i]
else:
number = list(str(int(''.join(number[:j]))+1))+number[j:]
else:
number[j] = number[i]
i += 1
j -= 1
# Concatenate and return the result.
return "".join(number)
This problem has a wonderful number of ways to solve them.
One of them is
def nearest_palindrome(number):
#start writitng your code here
while True:
number+=1
if str(number) == str(number)[::-1]:
return number
number=12300
print(nearest_palindrome(number))
Thanks for your time to read my answer : )
I have written this for finding next pallindrome number given a pallindrome number.
def palindrome(num):
bol=False
#x=len(str(num))
num=num+1
while(bol==False):
if(check_palindrome(num)):
bol=True
else:
num=num+1
return num
def check_palindrome(n):
temp=n
rev=0
while(n>0):
dig=n%10
rev=rev*10+dig
n=n//10
if(temp==rev):
return True
b=palindrome(8)
print(b)
def next_palin_drome(n):
while True:
n+=1
if str(n) == str(n)[::-1]:
return n
n=12231
print(next_palin_drome(n))
output:12321
def nearest_palindrome(number):
for i in range(1,number):
number=number+1
tem=str(number)
tem1=tem[-1::-1]
if(tem==tem1):
return number
else:
continue
number=12997979797979797
print(nearest_palindrome(number))
def nearest_palindrome(number):
n = len(str(number))//2
if(len(str(number)) % 2 == 0):
#number like 1221
number_1 = int((str(number))[:n]) #12
number_2 = int((str(number))[n:]) #21
if(number_1 < number_2):
number_1 += 1
number_2 = int(str(number_1)[::-1])
else:
number_2 = int(str(number_1)[::-1])
# if last half part is zero then just reverse the first number
if number_2 == 0:
number_2 = str(number_1)[::-1]
#combining the both parts
ans = int(str(number_1) + str(number_2))
return ans
else:
#numer like 12510 n=2
nu = int((str(number))[:n+1]) #add in this number
number_1 = int((str(number))[:n]) # 12
number_2 = int((str(number))[n+1:]) # 21
if (number_1 < number_2):
nu += 1
number_2 = int((str(nu))[::-1][1:])
else:
number_2 = int((str(nu))[::-1][1:])
#if last half part is zero then just reverse the first number
if number_2 == 0:
number_2 = str(nu)[::-1]
number_2 = number_2[1:]
#combinning both parts
ans = int(str(nu) + str(number_2))
return ans
number=12331
print(nearest_palindrome(number))
If a definite range is given:
# function to check if the number is a palindrome
def palin(x):
s=str(x)
if s==s[::-1]:
return True
else:
return False
n=int(input("Enter the number"))
# Putting up range from the next number till 15 digits
for i in range(n+1,int(10e14)):
if palin(i) is True:
print(i)
break
A brute force method:
def math(n):
while not var:
n += 1
if str(n) == str(n)[::-1] : f = 'but next is : '+str(n); return f
n = int(input()); t = math(n); print('Yes',t) if str(n) == str(n)[::-1] else print('No',t); global var; var = False
This is a good fast solution. I saw that the other solutions were iterating and checking through every +1 they did, but this is really slow for big numbers.
This solution has O(n) time if you look at the length of the number
beginNumber = 123456789101112131415161718 #insert number here for next palidrome
string = str(beginNumber + 1)
length = len(string)
number= [int(x) for x in list(string)]
for i in range(length//2):
if (number[i] != number[length-1-i]):
if (number[i]<number[length-1-i]):
number[length-2-i] += 1
number[length-1-i] = number[i]
print("".join([str(x) for x in number]))
I have written this for finding next pallindrome number given a pallindrome number..
#given a pallindrome number ..find next pallindrome number
input=999
inputstr=str(input)
inputstr=inputstr
#append 0 in beginning and end of string ..in case like 99 or 9999
inputstr='0'+inputstr+'0'
length=len(inputstr)
halflength=length/2;
#if even length
if(length%2==0):
#take left part and reverse it(which is equal as the right part )
temp=inputstr[:length/2]
temp=temp[::-1]
#take right part of the string ,move towards lsb from msb..If msb is 9 turn it to zero and move ahead
for j,i in enumerate(temp):
#if number is not 9 then increment it and end loop
if(i!="9"):
substi=int(i)+1
temp=temp[:j]+str(substi)+temp[j+1:]
break;
else:
temp=temp[:j]+"0"+temp[j+1:]
#now you have right hand side...mirror it and append left and right part
output=temp[::-1]+temp
#if the length is odd
if(length%2!=0 ):
#take the left part with the mid number(if length is 5 take 3 digits
temp=inputstr[:halflength+1]
#reverse it
temp=temp[::-1]
#apply same algoritm as in above
#if 9 then make it 0 and move on
#else increment number and break the loop
for j,i in enumerate(temp):
if(i!="9"):
substi=int(i)+1
temp=temp[:j]+str(substi)+temp[j+1:]
break;
else:
temp=temp[:j]+"0"+temp[j+1:]
#now the msb is the middle element so skip it and copy the rest
temp2=temp[1:]
#this is the right part mirror it to get left part then left+middle+right isoutput
temp2=temp2[::-1]
output=temp2+temp
print(output)
similarly for this problem take the left part of given number ...reverse it..store it in temp
inputstr=str(number)
if(inputstr==inputstr[::-1])
print("Pallindrome")
else:
temp=inputstr[:length/2]
temp=temp[::-1]
for j,i in enumerate(temp):
if(i!="9"):
substi=int(i)+1
temp=temp[:j]+str(substi)+temp[j+1:]
break;
else:
temp=temp[:j]+"0"+temp[j+1:]
now depending on length of your number odd or even generate the output..as in the code
if even then output=temp[::-1]+temp
if odd then temp2=temp1[1:]
output=temp2[::-1]+temp
I am not sure about this solution..but hope it helps