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Repeat string with a specific pattern without slicing

I'm working on a pretty annoying Python assignment and I'm lost. These are the requirements:
Ask for input.
For each character in the saved string I need to output said string with a certain modification. For example, if the input is abcd the output looks like this:
abcd
bcda
cdab
dabc
I.e. there are len(input) lines, each line begins with the next input[i] character and repeats to the length of the original input.
I should not use slicing, it's loop practice (T_T). No functions or packages. Loops only.
I made a working script that looks like this:
w = input('Type a word:')
w2 = ''
for i, char in enumerate(w):
w2 = w[i:]+w[:i]
print(w2)
It's neat and short. But it will be marked down for slicing. Can Python loop gurus please help me remake it into loops? Thanks so much in advance!

You can use indexing into the original string using a modulo on itself:
w = "aword"
lw = len(w)
for offset in range(lw):
for character in range(lw):
print(w[(offset+character) % lw], end="")
print()
Output:
aword
worda
ordaw
rdawo
dawor
If your sum of offset and character overshoots the amount of characters the modulo operation wraps it around.

If you can't slice strings, you can append and pop lists. So, convert the string to a list and work with list methods.
>>> test = "abcd"
>>> l = list(test)
>>> for _ in range(len(l)):
... print("".join(l))
... l.append(l.pop(0))
...
abcd
bcda
cdab
dabc

Just for fun another one:
s = 'aword'
ss = s * 2 # 'awordaword'
for i in range(len(s)):
for j in range(len(s)):
print(ss[i+j], end='')
print()
Output:
aword
worda
ordaw
rdawo
dawor

How to get rid of a repeating word in this function?

I don't know what kind of problem this is called and couldn't find something related to this on the site. I am trying to make a function that prints out the first letter of the word on the first line then the next two letters and so on. However, I am not sure on how to prevent repeating the complete word. For example in the word 'bar', it should go
b
ba
bar
ba
b
but my function repeats bar twice. Thanks!
a= []
def letters():
x = input("Enter a string")
count = 0
for c in x:
count +=1
y = 0
while y <= count:
z = (x[:y])
a.append(z)
y += 1
negWords = a[::-1]
for words in a:
print (words)
for words in negWords:
print (words)

You seem to be doing some unnecessary work for simply wanting to print something. You can make use of len to get the length of the word, without having to go through a loop to get the size.
Also, collecting the data in a list seems unnecessary.
To stick to the nature of your approach with the while loop, you can go through the entire length of the word and back again until "0". So, to me, I see that as twice the length of the word. With that in mind, that would be my control for the while loop. I would simply then check for when my incrementer reaches about the length of the word, and start going backwards from there:
w = "bar"
l = len(w)
i = 0
while i <= l*2:
if i > l:
print(w[:(l - i)])
else:
print(w[:i])
i += 1
Output:
b
ba
bar
ba
b

Try this:
negWords= a[:-1:-1]
This will not include the full word a second time
Also, your code could use a lot of cleanup. For example:
count = len(x)
Makes more sense than what you have

How to strip white spaces in Python without using a string method?

I am fairly new to programming and have been learning some of the material through HackerRank. However, there is this one objective or challenge that I am currently stuck on. I've tried several things but still cannot figure out what exactly I am doing wrong.
Objective: Read N and output the numbers between 0 and N without any white spaces or using a string method.
N = int(input())
listofnum = []
for i in range(1, N +1):
listofnum.append(i)
print (*(listofnum))
Output :
1 2 3

N = int(input())
answer = ''
for i in range(1, N + 1):
answer += str(i)
print(answer)
This is the closest I can think of to 'not using any string methods', although technically it is using str.__new__/__init__/__add__ in the background or some equivalent. I certainly think it fits the requirements of the question better than using ''.join.

Without using any string method, just using integer division and list to reverse the digits, print them using sys.stdout.write:
import sys
N = int(input())
for i in range(1,N+1):
l=[]
while(i):
l.append(i%10)
i //= 10
for c in reversed(l):
sys.stdout.write(chr(c+48))
Or as tdelaney suggested, an even more hard-code method:
import os,sys,struct
N = int(input())
for i in range(1,N+1):
l=[]
while(i):
l.append(i%10)
i //= 10
for c in reversed(l):
os.write(sys.stdout.fileno(), struct.pack('b', c+48))
All of this is great fun, but the best way, though, would be with a one-liner with a generator comprehension to do that, using str.join() and str construction:
"".join(str(x) for x in range(1,N+1))
Each number is converted into string, and the join operator just concatenates all the digits with empty separator.

You can print numbers inside the loop. Just use end keyword in print:
print(i, end="")

Try ''.join([str(i) for i in range(N)])

One way to accomplish this is to append the numbers to a blank string.
out = ''
for i in range(N):
out += str(i)
print(out)

You can make use of print()'s sep argument to "bind" each number together from a list comprehension:
>>> print(*[el for el in range(0, int(input())+1)], sep="")
10
012345678910
>>>

You have to do a simple math to do this. What they expect to do is multiply each of your list elements by powers of ten and add them up on each other. As an example let's say you have an array;
a = [2,3,5]
and you need to output;
235
Then you multiply each of loop elements starting from right to left by 10^0, 10^1 and 10^2. You this code after you make the string list.
a = map(int,a)
for i in range(len(a)):
sum += (10**i)*a[-i]
print sum
You are done!

create and insert randomly duplicates

This is not probably a common question but I want to CREATE duplicates and INSERT it randomly in my String. Just like the following example.
I have this file :
AppleStrawberryBananaCitrusOrange
And I expected this kind of output :
trusppleStrawbeApplertrusryBananaCitrusOrangepple
In this case my program randomly select a substring of length '4' : 'pple' and 'trus' and duplicates him 'twice(2)' before the insertion.
I think that I could run the program by using the fonction copy with copy.copy() and copy.insert() but I don't really know how to use it randomly.
For the moment;I just write the part of the code for read and write and something else:
import copy
chain='*'
contain = ''
file= raw_input ('Filename:')
x = open(file,'r')
for line in x:
if not(chain in line):
contain+=line
e=copy.copy(contain[4:8])
f=copy.copy(contain[8:12])
y = open('copi','w')
y.write(contain)
y.write(f)
x.close()
Result:
AppleStrawberryBananaCitrusOrange
awbe
As you can see; it doesn't really work like I want. :(
Thanks for your help

Not sure if I understand what you are trying to do.
You'll probably need a library for random selection:
import random
Now here is your input string:
s = "AppleStrawberryBananaCitrusOrange"
print s
You can use random.randint to select a random position and extract a 4-character word:
i = random.randint(0, len(s) - 4)
w = s[i:i+4]
print w
Finally, you can select two more random positions and insert the word via string concatenation:
for j in range(2):
i = random.randint(0, len(s) - 4)
s = s[:i] + w + s[i:]
print s
Output:
AppleStrawberryBananaCitrusOrange
yBan
ApyBanpleSyBantrawberryBananaCitrusOrange

Reverse a string without using reversed() or [::-1]?

I came across a strange Codecademy exercise that required a function that would take a string as input and return it in reverse order. The only problem was you could not use the reversed method or the common answer here on stackoverflow, [::-1].
Obviously in the real world of programming, one would most likely go with the extended slice method, or even using the reversed function but perhaps there is some case where this would not work?
I present a solution below in Q&A style, in case it is helpful for people in the future.

You can also do it with recursion:
def reverse(text):
if len(text) <= 1:
return text
return reverse(text[1:]) + text[0]
And a simple example for the string hello:
reverse(hello)
= reverse(ello) + h # The recursive step
= reverse(llo) + e + h
= reverse(lo) + l + e + h
= reverse(o) + l + l + e + h # Base case
= o + l + l + e + h
= olleh

Just another option:
from collections import deque
def reverse(iterable):
d = deque()
d.extendleft(iterable)
return ''.join(d)

Use reversed range:
def reverse(strs):
for i in xrange(len(strs)-1, -1, -1):
yield strs[i]
...
>>> ''.join(reverse('hello'))
'olleh'
xrange or range with -1 step would return items in reversed order, so we need to iterate from len(string)-1 to -1(exclusive) and fetch items from the string one by one.
>>> list(xrange(len(strs) -1, -1 , -1))
[4, 3, 2, 1, 0] #iterate over these indexes and fetch the items from the string
One-liner:
def reverse(strs):
return ''.join([strs[i] for i in xrange(len(strs)-1, -1, -1)])
...
>>> reverse('hello')
'olleh'

EDIT
Recent activity on this question caused me to look back and change my solution to a quick one-liner using a generator:
rev = ''.join([text[len(text) - count] for count in xrange(1,len(text)+1)])
Although obviously there are some better answers here like a negative step in the range or xrange function. The following is my original solution:
Here is my solution, I'll explain it step by step
def reverse(text):
lst = []
count = 1
for i in range(0,len(text)):
lst.append(text[len(text)-count])
count += 1
lst = ''.join(lst)
return lst
print reverse('hello')
First, we have to pass a parameter to the function, in this case text.
Next, I set an empty list, named lst to use later. (I actually didn't know I'd need the list until I got to the for loop, you'll see why it's necessary in a second.)
The count variable will make sense once I get into the for loop
So let's take a look at a basic version of what we are trying to accomplish:
It makes sense that appending the last character to the list would start the reverse order. For example:
>>lst = []
>>word = 'foo'
>>lst.append(word[2])
>>print lst
['o']
But in order to continue reversing the order, we need to then append word[1] and then word[0]:
>>lst.append(word[2])
>>lst.append(word[1])
>>lst.append(word[0])
>>print lst
['o','o','f']
This is great, we now have a list that has our original word in reverse order and it can be converted back into a string by using .join(). But there's a problem. This works for the word foo, it even works for any word that has a length of 3 characters. But what about a word with 5 characters? Or 10 characters? Now it won't work. What if there was a way we could dynamically change the index we append so that any word will be returned in reverse order?
Enter for loop.
for i in range(0,len(text)):
lst.append(text[len(text)-count])
count += 1
First off, it is necessary to use in range() rather than just in, because we need to iterate through the characters in the word, but we also need to pull the index value of the word so that we change the order.
The first part of the body of our for loop should look familiar. Its very similar to
>>lst.append(word[..index..])
In fact, the base concept of it is exactly the same:
>>lst.append(text[..index..])
So what's all the stuff in the middle doing?
Well, we need to first append the index of the last letter to our list, which is the length of the word, text, -1. From now on we'll refer to it as l(t) -1
>>lst.append(text[len(text)-1])
That alone will always get the last letter of our word, and append it to lst, regardless of the length of the word. But now that we have the last letter, which is l(t) - 1, we need the second to last letter, which is l(t) - 2, and so on, until there are no more characters to append to the list. Remember our count variable from above? That will come in handy. By using a for loop, we can increment the value of count by 1 through each iteration, so that the value we subtract by increases, until the for loop has iterated through the entire word:
>>for i in range(0,len(text)):
..
.. lst.append(text[len(text)-count])
.. count += 1
Now that we have the heart of our function, let's look at what we have so far:
def reverse(text):
lst = []
count = 1
for i in range(0,len(text)):
lst.append(text[len(text)-count])
count += 1
We're almost done! Right now, if we were to call our function with the word 'hello', we would get a list that looks like:
['o','l','l','e','h']
We don't want a list, we want a string. We can use .join for that:
def reverse(text):
lst = []
count = 1
for i in range(0,len(text)):
lst.append(text[len(text)-count])
count += 1
lst = ''.join(lst) # join the letters together without a space
return lst
And that's it. If we call the word 'hello' on reverse(), we'd get this:
>>print reverse('hello')
olleh
Obviously, this is way more code than is necessary in a real life situation. Using the reversed function or extended slice would be the optimal way to accomplish this task, but maybe there is some instance when it would not work, and you would need this. Either way, I figured I'd share it for anyone who would be interested.
If you guys have any other ideas, I'd love to hear them!

Only been coding Python for a few days, but I feel like this was a fairly clean solution. Create an empty list, loop through each letter in the string and append it to the front of the list, return the joined list as a string.
def reverse(text):
backwardstext = []
for letter in text:
backwardstext.insert(0, letter)
return ''.join(backwardstext)

I used this:
def reverse(text):
s=""
l=len(text)
for i in range(l):
s+=text[l-1-i]
return s

Inspired by Jon's answer, how about this one
word = 'hello'
q = deque(word)
''.join(q.pop() for _ in range(len(word)))

This is a very interesting question, I will like to offer a simple one
liner answer:
>>> S='abcdefg'
>>> ''.join(item[1] for item in sorted(enumerate(S), reverse=True))
'gfedcba'
Brief explanation:
enumerate() returns [(0, 'a'), (1, 'b'), (2, 'c'), (3, 'd'), (4, 'e'), (5, 'f'), (6, 'g')]. The indices and the values.
To reverse the values, just reverse sort it by sorted().
Finally, just put it together back to a str

I created different versions of how to reverse a string in python in my repo:
https://github.com/fedmich/Python-Codes/tree/master/Reverse%20a%20String
You can do it by using list-comprehension or lambda technique:
# Reverse a string without using reverse() function
s = 'Federico';
li = list( s ) #convert string to list
ret = [ li[i-1] for i in xrange(len(li),0,-1) ] #1 liner lambda
print ( "".join( ret ) )
or by doing a backward for loop
# Reverse a string without using reverse() function
s = 'Federico';
r = []
length = len(s)
for i in xrange(length,0,-1):
r.append( s[ i - 1] )
print ( "".join(r) )

reduce(lambda x, y : y + x, "hello world")

A golfed version: r=lambda x:"".join(x[i] for i in range(len(x-1),-1,-1)).

i just solved this in code academy and was checking my answers and ran across this list. so with a very limited understanding of python i just did this and it seamed to work.
def reverse(s):
i = len(s) - 1
sNew = ''
while i >= 0:
sNew = sNew + str(s[i])
i = i -1
return sNew

def reverse(s):
return "".join(s[i] for i in range(len(s)-1, -1, -1))

Blender's answer is lovely, but for a very long string, it will result in a whopping RuntimeError: maximum recursion depth exceeded. One might refactor the same code into a while loop, as one frequently must do with recursion in python. Obviously still bad due to time and memory issues, but at least will not error.
def reverse(text):
answer = ""
while text:
answer = text[0] + answer
text = text[1:]
return answer

Today I was asked this same exercise on pen&paper, so I come up with this function for lists:
def rev(s):
l = len(s)
for i,j in zip(range(l-1, 0, -1), range(l//2)):
s[i], s[j] = s[j], s[i]
return s
which can be used with strings with "".join(rev(list("hello")))

This is a way to do it with a while loop:
def reverse(s):
t = -1
s2 = ''
while abs(t) < len(s) + 1:
s2 = s2 + s[t]
t = t - 1
return s2

I have also just solved the coresponding exercise on codeacademy and wanted to compare my approach to others. I have not found the solution I used so far, so I thought that I sign up here and provide my solution to others. And maybe I get a suggestion or a helpful comment on how to improve the code.
Ok here it goes, I did not use any list to store the string, instead I have just accessed the string index. It took me a bit at first to deal with the len() and index number, but in the end it worked :).
def reverse(x):
reversestring = ""
for n in range(len(str(x))-1,-1, -1):
reversestring += x[n]
return reversestring
I am still wondering if the reversestring = "" could be solved in a more elegant way, or if it is "bad style" even, but i couldn't find an answer so far.

def reverse(text):
a=""
l=len(text)
while(l>=1):
a+=text[l-1]
l-=1
return a
i just concatenated the string a with highest indexes of text (which keeps on decrementing by 1 each loop).

All I did to achieve a reverse string is use the xrange function with the length of the string in a for loop and step back per the following:
myString = "ABC"
for index in xrange(len(myString),-1):
print index
My output is "CBA"

You can simply reverse iterate your string starting from the last character. With python you can use list comprehension to construct the list of characters in reverse order and then join them to get the reversed string in a one-liner:
def reverse(s):
return "".join([s[-i-1] for i in xrange(len(s))])
if you are not allowed to even use negative indexing you should replace s[-i-1] with s[len(s)-i-1]

You've received a lot of alternative answers, but just to add another simple solution -- the first thing that came to mind something like this:
def reverse(text):
reversed_text = ""
for n in range(len(text)):
reversed_text += text[-1 - n]
return reversed_text
It's not as fast as some of the other options people have mentioned(or built in methods), but easy to follow as we're simply using the length of the text string to concatenate one character at a time by slicing from the end toward the front.

def reverseThatString(theString):
reversedString = ""
lenOfString = len(theString)
for i,j in enumerate(theString):
lenOfString -= 1
reversedString += theString[lenOfString]
return reversedString

This is my solution using the for i in range loop:
def reverse(string):
tmp = ""
for i in range(1,len(string)+1):
tmp += string[len(string)-i]
return tmp
It's pretty easy to understand. I start from 1 to avoid index out of bound.

Here's my contribution:
def rev(test):
test = list(test)
i = len(test)-1
result = []
print test
while i >= 0:
result.append(test.pop(i))
i -= 1
return "".join(result)

You can do simply like this
def rev(str):
rev = ""
for i in range(0,len(str)):
rev = rev + str[(len(str)-1)-i]
return rev

Here is one using a list as a stack:
def reverse(s):
rev = [_t for _t in s]
t = ''
while len(rev) != 0:
t+=rev.pop()
return t

Try this simple and elegant code.
my_string= "sentence"
new_str = ""
for i in my_string:
new_str = i + new_str
print(new_str)

you have got enough answer.
Just want to share another way.
you can write a two small function for reverse and compare the function output with the given string
var = ''
def reverse(data):
for i in data:
var = i + var
return var
if not var == data :
print "No palindrome"
else :
print "Palindrome"

Not very clever, but tricky solution
def reverse(t):
for j in range(len(t) // 2):
t = t[:j] + t[- j - 1] + t[j + 1:- j - 1] + t[j] + t[len(t) - j:]
return t

Pointfree:
from functools import partial
from operator import add
flip = lambda f: lambda x, y: f(y, x)
rev = partial(reduce, flip(add))
Test:
>>> rev('hello')
'olleh'