I've written a code which works pretty well, no errors, no problems. But no matter how hard I try to print the list values each in a new line, it still wouldn't work. I've tried sep='\n', and even tried to put the list in a loop to print each value one by one. I still get the result printed all in a row, in one line. This sounds too annoying and I can't figure out why my code is having this strange behavior. Here's my code:
length = int(input())
input_string = [int(y) for y in input().split()]
def lowest(string):
return(min(x for x in string if x is not None))
def none_set(string):
for k in range(length):
if string[k] != None:
if string[k] <=0:
string[k] = None
def counter(string):
nums = 0
for h in range(length):
if string[h] != None:
nums += 1
return nums
cnt = []
for i in range(length):
minimum = lowest(input_string)
none_set(input_string)
cnt.append(counter(input_string))
for j in range(length):
if input_string[j] != None:
input_string[j] -= minimum
result = list(set(cnt))[::-1]
print(result, sep='\n') #Doesn't print the values in new line :/
Sample Input:
6
5 4 4 2 2 8
Expected Output:
6
4
2
1
The Output I Get:
[6, 4, 2, 1]
In case you want to know what exactly my code does, check this link here (No login/signup needed), however, the issue is not really relative to what the goal of my code is, I'd say.
I appreciate in advance, for any tip, solution, or help.
It's because you're printing the whole list.
lst = ['a','b','c']
If I print this list I get ['a','b','c']. To print each item you can use a for loop like so:
lst = ['a','b','c']
for item in lst:
print(item)
#output:
a
b
c
Try this.
for i in result:
print(i)
Try to print like that:
for i in result:
print(i)
Use join() instead of multiple calls to print(). Some will consider it more difficult to read, but it's definitely more efficient even for small lists, and orders of magnitude faster for lists greater than 100 elements.
print("\n".join(map(str, result)))
I have been trying to use lists so I can separate my integers into something organized (which lists are made to do, right?). Making these integers as random numbers are also makes this a little more confusing. This is what I have so far:
import random
first_list = []
for first_num in range(5):
first_list = (random.randint(1,9))
print ("First List",first_list)
for x in range(5):
num_two = random.randint(2,8)
print ("Second List",num_two)
My display is random numbers going down with my printed First List/Second List connected to every number.
First List:8
First List 3
etc.
Second List:3
Second List 7
etc.
Second part of my assignment is to use to compare elements in the two lists in pairs, i.e., compare the first elements in both lists, compare the second elements in the both lists which has to show the larger number in each comparison. I just wanted to show what the conclusion is as to why I need help from you all in the first place.
Your professor/teacher/whoever is testing your knowledge of the list.append() method and the zip() method.
list.append() adds an element to a list. zip() yields an object that allows for iterating through multiple iterables at the same time.
read the python docs.
https://docs.python.org/3/
import random
first_list = []
second_list = []
for x in range(5):
first_list.append(random.randint(1,9))
for y in range(5):
second_list.append(random.randint(1,9))
print('First List:\n')
for num in first_list:
print(str(num)+',\n')
print('Second List:\n')
for num in second_list:
print(str(num)+',\n')
both_lists = zip(first_list, second_list)
for first, second in both_lists:
if first > second:
print('First is greater: ' + str(first))
elif first < second:
print('Second is greater: ' + str(second))
else:
print('First and second are equal: ' + str(first) + ', ' + str(second))
Martijn's answer is much more pythonic and elegant, although for printing the results, max() would need to be extended with a custom function to pass to map(), but I'm assuming that you don't understand list comprehension. This is the beginner's answer.
Your question is confusing, May be you are looking for something like this?
import random
first_list = []
Second_list=[]
for first_num in range(5):
first_list.append(random.randint(1,9))
for x in range(5):
Second_list.append(random.randint(2,8))
print([max(i) for i in zip(first_list,Second_list)])
You can also try :
print(list(map(max,zip(first_list,Second_list))))
Shorter version of your code:
import random
first_list = [random.randint(1,9) for first_num in range(5)]
Second_list=[random.randint(2,8) for x in range(5)]
print(list(map(max,zip(first_list,Second_list))))
a=[1,2,3]
print "the list is ：%"%(a)
I want to print out one line like this: the list is：[1,2,3]
I can not make it within one line and I have to do in this way:
print " the list is :%"
print a
I am wondering whether I can print out something that combine with string formatting and a list in ONE line.
The simplest way is what CodeHard_or_HardCode said in the comments. For Python 3 it would be:
a=[1,2,3]
print('This is a list', a)
This is a list [1, 2, 3]
I suggest using the string format method, as it is recommended over the old % syntax.
print("the list is: {}".format(a))
At least in Python 2.7.4., this will work:
print " the list is " + str(a)
Try this:
a = [1,2,3]
print("the list is: %s" % a)
print(f"the list is: {[1,2,3]}")
you will need to use python 3.7.1 or higher I believe, as this is when they added string interpolation
a = [1,2,3]
print(f'the list is: {a}')
I have been trying to print the results of a function that happen to be in a list. If this doesn't make sense.
In print_long.py, define a Python function print_long(list) that takes a list of strings called list and prints out each string in list that has a length greater than or equal to 6.
That is the original question. I tried doing
def print_long(list):
for i in print_long(list):
if len(list) > 6:
print (i)
else:
return None
but that is clearly wrong since it infinitely loops it. Or it seems to appear to do that.
help.
Your code appears to be a recursive call with no base (terminating) condition. That is, when you first call print_long() you are immediately calling print_long(), which calls print_long() ad infinitum.
The code you were probably attempting is something like:
def greater_than(mylist, n):
for item in mylist:
if len(item) > n:
print item
mylist = ["hello", "foo", "bar", "testing", "world"]
greater_than(mylist, 4) # Prints: hello, testing, world
Or, more simply:
mylist = ["hello", . . ., "world"] # Same as above
n = 4
[x for x in mylist if len(x) > n] # ['hello', 'testing', 'world']
I think this is what you are trying to do:
def print_long(in_list): # define your function
for i in in_list: # loop over the input list
if len(i) > 6: # test the length of the elements
print (i) # if true print it out
else: # you don't need this else, as you really just want
pass # to do nothing if len(i) < 6
Name your parameter something other than list. Jonathon Reinhart gave a good suggestion: mylist.
The other code examples above also provide good suggestions for writing your for loop better:
for i in mylist:
You might also need to change the len() to => 6.
if len(i) >= 6
From what you wrote, it look says, "prints out each string in list that has a length greater than or equal to 6." So you'll need the equal to make sure you capture lengths equal to 6
A short one:
def print_greater_or_equal_than(mylist, n):
print '\n'.join(filter(lambda i: len(i) >= n, mylist))
You are making a recursive call to your function without reducing your list input, first. Therefore, it will continue to make recursive calls to infinity.
To avoid the infinite stack of recursion, print the item in the list & remove it before recursing :
def print_long(list):
i = list.pop(0)
if len(list) > 6:
print (i)
else:
return None
if len(list):
print_long(list)
First, list is a reserved keyword in Python, so don't use it as a parameter or variable name. I usually use lst. Anyway, there are a few other issues to work out here.
def print_long(lst):
for i in lst:
if len(i) >= 6:
print i
print_long(['foo','bar','hello','johnny'])
Before, you were trying to iterate through your function:
for i in print_long(lst)
This is incorrect, you simply want to iterate through the list, as shown in the first example above. A more pythonic way of achieving the result you are looking for is:
print ' '.join([i for i in lst if len(i) >= 6])
Look up Python list comprehensions if you don't understand.
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'