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).
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.
The task is:
User enters a number, you take 1 number from the left, one from the right and sum it. Then you take the rest of this number and sum every digit in it. then you get two answers. You have to sort them from biggest to lowest and make them into a one solid number. I solved it, but i don't like how it looks like. i mean the task is pretty simple but my code looks like trash. Maybe i should use some more built-in functions and libraries. If so, could you please advise me some? Thank you
a = int(input())
b = [int(i) for i in str(a)]
closesum = 0
d = []
e = ""
farsum = b[0] + b[-1]
print(farsum)
b.pop(0)
b.pop(-1)
print(b)
for i in b:
closesum += i
print(closesum)
d.append(int(closesum))
d.append(int(farsum))
print(d)
for i in sorted(d, reverse = True):
e += str(i)
print(int(e))
input()
You can use reduce
from functools import reduce
a = [0,1,2,3,4,5,6,7,8,9]
print(reduce(lambda x, y: x + y, a))
# 45
and you can just pass in a shortened list instead of poping elements: b[1:-1]
The first two lines:
str_input = input() # input will always read strings
num_list = [int(i) for i in str_input]
the for loop at the end is useless and there is no need to sort only 2 elements. You can just use a simple if..else condition to print what you want.
You don't need a loop to sum a slice of a list. You can also use join to concatenate a list of strings without looping. This implementation converts to string before sorting (the result would be the same). You could convert to string after sorting using map(str,...)
farsum = b[0] + b[-1]
closesum = sum(b[1:-2])
"".join(sorted((str(farsum),str(closesum)),reverse=True))
I am new to coding and is trying to solve this python question
Question:
Write a program that calculates and prints the value according to the given formula:
Q = Square root of [(2 * C * D)/H]
Following are the fixed values of C and H:
C is 50. H is 30.
D is the variable whose values should be input to your program in a comma-separated sequence.
Example
Let us assume the following comma separated input sequence is given to the program:
100,150,180
The output of the program should be:
18,22,24
Hints:
If the output received is in decimal form, it should be rounded off to its nearest value (for example, if the output received is 26.0, it should be printed as 26)
In case of input data being supplied to the question, it should be assumed to be a console input.
This is the solution given. I have not seen 'x for x in input()'expression, may I know what does this expression do ?
import math
c=50
h=30
value = []
items=[x for x in input().split(',')]
for d in items:
value.append(str(int(round(math.sqrt(2*c*float(d)/h)))))
print (','.join(value))
This is my own solution but somehow I got a syntax error.
def sr(D):
For item in D:
return ((2*50*D)/30)**0.5
try:
a=int(input())
j=a.split(",")
print(sr(j))
except:
print('Please enter an integers or intergers seperated by comma')
The x is just a variable that gets assigned to the input that comes in via the input() function.
If you're aware of C style language (or Java), it's similar to
for(int i=0;<some_condition>;<some_operation>){}
This is just a condensed, pythonic and easy to read way to do this.
You can read more Python loops here
https://wiki.python.org/moin/ForLoop
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