How can I do this effectively? [closed] - python

Closed. This question needs details or clarity. It is not currently accepting answers.
Want to improve this question? Add details and clarify the problem by editing this post.
Closed 2 years ago.
Improve this question
This problem:
Input: 123456
Result:
1+2+3+4+5+6 = 21
2+1 = 3
Return: 3
This is my code:
num = input()
print(sum(list(map(int, list(num)))))
I don't know how to do until it is 1 digit.

Try this (example in IPython):
In [1]: s = '123456'
Out[1]: '123456'
In [2]: digits = [int(j) for j in s]
Out[2]: [1, 2, 3, 4, 5, 6]
In [3]: s = str(sum(digits))
Out[3]: '21'
Repeat steps 2 and three until len(s) == 1.

One way:
while len(str(ans))>1:
ans = sum(map(int, str(ans)))
Full code:
num = 45637
ans = num
while len(str(ans))>1:
ans = sum(map(int, str(ans)))
print(ans)
Output for input 45637:
7

You can try this:
s = input()
while(len(s)>1):
s = str(sum(list(map(int,s))))

One way to do it using sum(), list comprehension and recursion,
def simulated_sum(input):
"""This is a recursive function
to find the simulated sum of an integer"""
if len(str(input)) == 1:
return input
else:
input_digits = [int(x) for x in str(input)]
latest_sum = sum(input_digits)
return simulated_sum(latest_sum)
input = int(input('Enter a number'))
print(simulated_sum(input))
DEMO: https://rextester.com/WCBXIL71483

Is this what you want? (instructions unclear):
def myfunction(number):
total = 0
answertotal = 0
for i in str(number):
total += int(i)
for i in str(total):
answertotal += int(i)
return answertotal
myfunction(123456)
This function returns 3

Related

can you explain how this Loop works and how do we get that output? [closed]

Closed. This question needs details or clarity. It is not currently accepting answers.
Want to improve this question? Add details and clarify the problem by editing this post.
Closed 1 year ago.
Improve this question
if __name__ == '__main__':
n = int(input())
for i in range (0,n):
result = i**2
print(result)
#input : 5
#output : 0 1 4 9 16
range 0,n = 0,1,2,3,4 and
we gave i**2 but how we got 0,1,4,9,16 as output?
range(start, stop, step)
** = square of Number
start Optional. An integer number specifying at which position to
start. Default is 0
stop Required. An integer number specifying at which position to
stop (not included).
step Optional. An integer number specifying the incrementation. Default is 1
you are passing required parameter as 5 which will not be included in the loop. so as per your calculation it will start from 0
result = 0**2 = 0
result = 1**2 = 1
result = 2**2 = 4
result = 3**2 = 9
result = 4**2 = 16
'i' will not reach 5 because of non-inclusive nature of range() operator.

In Python, how can I make it so the first 2 elements of a list are added together, then the first 4, then the first 6, and so on? [closed]

Closed. This question needs to be more focused. It is not currently accepting answers.
Want to improve this question? Update the question so it focuses on one problem only by editing this post.
Closed 2 years ago.
Improve this question
I am making a game where two players take turns adding a number 1 to 10 to the running total. The player who adds to get 100 wins. I stored the inputs of the players in a list, and at the end of the game, I want to display the total after every turn. The list is in the format:
allTurns = [player1move1, player2move1, player1move2, player2move2, ...]
How can I add only the first 2 elements and display them, then add only the first 4 elements and display them, and then the first 6, and so on?
for i in range(0, len(list)):
if i % 2 == 0:
print(sum(list[:i+1]))
Or
sums = []
for i in range(0, len(list)):
if i % 2 == 0:
sums.append(sums[-1] + list[i-1] + list[i])
Try this:
allTurns = [1,2,3,6,12,23]
out = []
for n in range(len(allTurns)):
if n % 2 is 1:
out.append(allTurns[n] + allTurns[n-1])
if len(allTurns) % 2 is 1:
out.append(allTurns[-1])
print(out)
I've coded it so that it could work for any situation, and not just yours.
Try this:
allTurns = [1,2,3,6,12,23,4]
out = []
for n in allTurns:
if allTurns.index(n) % 2 is 0:
i = 0
for a in range(allTurns.index(n)+2):
try:
i=i+allTurns[a]
except IndexError:
pass
out.append(i)
print(out)
This is probably as simple as it can get.

Extract n first consecutive numbers from a string in python [closed]

Closed. This question needs to be more focused. It is not currently accepting answers.
Want to improve this question? Update the question so it focuses on one problem only by editing this post.
Closed 2 years ago.
Improve this question
Let's consider this string :
st = 'text1text6253text'
How could I please extract the two first consecutive figures ?
Expected output :
62
You can either use regex with \d{2}and return that, or go over the string:
st = 'text1text6253text'
for i in range(len(st)-1):
if st[i].isdigit() and st[i+1].isdigit():
print(st[i]+st[i+1])
break
import re
def find_con(n, s):
result = re.search('\d{%s}'%n, s)
return result.group(0) if result else result
st = 'text1text6253text'
print(find_con(2, st))
st = 'text1text6253text'
lst = list(st)
lst2 = []
for i,v in enumerate(lst):
if lst[i].isdigit() and lst[i+1].isdigit():
lst2.append(lst[i])
lst2.append(lst[i+1])
ans = int(lst2[0] + lst2[1])
print(ans)
Thanks to your answers, I built a general function that I propose you below :
def extract_n_consecutive_numbers(st,nb):
for i in range(len(st)-nb+1):
is_numeric = True
for j in range(nb):
is_numeric = is_numeric & st[i+j].isdigit()
if is_numeric :
output = ""
for j in range(nb):
output += st[i+j]
return output
return ""
Example :
extract_n_consecutive_numbers('text1text6253text',2)
Out[1]: 62

How can I print the string "aaabbbccaa" as a3b3c2a2 [closed]

Closed. This question needs details or clarity. It is not currently accepting answers.
Want to improve this question? Add details and clarify the problem by editing this post.
Closed 2 years ago.
Improve this question
I am given a user entered string 'aaabbbccaa'.
I want to find the duplicates and print the string back as 'a3b3c2a2'
Maybe with this way:
from itertools import groupby
s = "aaabbbccaa"
# group by characters
groups = groupby(s)
# process result
result = "".join([label + str(len(list(group))) for label, group in groups])
print(result)
Output:
a3b3c2a2
def process_string(source):
new = ''
while source:
counter = 0
first_char = source[0]
while source and source[0] == first_char:
counter += 1
source = source[1:]
new += f'{first_char}{counter}'
return new
print(process_string('aaabbbccaa'))
'a3b3c2a2'
this kind of solution could mabye solve it, it does what you specify, however if you are able to put it into your context, no idea :)
hope it helps!
c = 0
foo = "aaabbbccaa"
bar = ""
prev = None
for counter, index in enumerate(foo):
print(c)
if prev == None:
print("first")
elif prev == index:
print("second")
elif prev != index:
c = 0
c += 1
prev = index
try:
if index in foo[counter+1]:
print("third")
else:
print("fourth")
bar += index + str(c)
except:
print("fifth")
bar += index + str(c)
print("foo is {}".format(foo)) # will output aaabbbccaa
print("bar is {}".format(bar)) # will output a3b3c2a2

is there possible to resolve this star pyramid [closed]

Closed. This question needs details or clarity. It is not currently accepting answers.
Want to improve this question? Add details and clarify the problem by editing this post.
Closed 8 years ago.
Improve this question
Today, i have trying to resolve a small star pyramid :
Input:
5 1
Output:
*
**
***
****
Code:
x = 1
y = 0
m, e = map(int, raw_input().split())
while x < m:
print "\n" * y, "*" * e
m -= 1
e += 1
I did that but there is a better solution?? Thanks =)
I think this can be solved more easily:
stop, first = map(int, raw_input().split())
for i in range(stop - 1):
print '*' * (i + first)
just for fun >:)
class c:
def __init__(s,m,e):
s.e , s.m = sorted([e, m])
s.r = 42
def __iter__(s):
return s
def next(s):
if s.m < s.e:
t = "".join(chr(s.r) for _ in range(s.m))
s.m += 1
return t
else:
raise StopIteration
print "\n".join(c(*map(int,raw_input().split())))
n = int(raw_input())
for i in range(n): print "*"*i
This appears to do what your program intends to do, however I can't quite tell because of the issues I raised in my comment above.

Categories

Resources