Your program should ask the user to input the list, then it should call the function, and print the result. with the following condition:
It returns True if the given list has at least 2 white spaces and False otherwise.
My code:
n = ((input("Please input a list of numbers separated by space:")))
t = 0
k = n.count(' ')
for i in range(0,len(n)):
if n[i] > " ":
print("True")
else:
print("False")
print("There are",k,"space which has two length run")
My program counts all of the white spaces, but I want it to only count the 2 white spaces or more and return True or False otherwise
I see several issues with this code.
First, you do not define a function as stated in the requirements.
Second, k = n.count(' ') already counts all the spaces in the input string. You do not need to loop over each character.
Third, even if you did need to loop over the characters, n[i] > " " is definitely not the right way to do what you want.
Fourth, what is the purpose of the t variable?
Fifth, why is input() enclosed in two extra layers of parentheses?
Not clear why you think you need a loop.
Try simply
print(n.count(' ') >= 2)
Or rather
def foo(n):
return n.count(' ') >= 2
print(foo(input("Enter some numbers:")))
Related
Second last line has the "problem".
This is the code:
cities = []
prompt = "Please enter the names of any five cities that you have visited."
prompt += "\nWe will compile them into a list for you"
num = 1
flag = True
while num <= 5 and flag:
city = input(f"City {num}: ")
if city.lower() == "quit":
break
else:
cities.append(city)
num += 1
print()
num = 0
place = ""
comp = ""
print("So you have visited the following cities:")
for place in cities:
num += 1
comp += f"{num}. {place}\t"
print(comp)
It is supposed to ask for 5 inputs and then number them and put them all in one sentence, as a string, then print that string.
It "works", there are no errors, but the space created in the result due to \t in the second last line always seems to vary. Sometimes there is no space at all, and sometimes there are two spaces. Same code, run over and over in command prompt, but different results every time in the behavior of \t.
It works perfectly if I replace the \t with four spaces.
What is the reason for this?
The tab just moves to the next tab stop. A tab does not equal four spaces. This is used to line up columns when printing multiple lines.
print("test\ttest")
print("test12\ttest")
print("test12345\ttest")
Output
test test
test12 test
test12345 test
Do you want to try to change your format statement to be something like this.
for place in cities:
num += 1
comp += f"{num}. {place:15}"
#your string length will always be left justified to 15 chars
print(comp)
string = input()
for i in range(len(string)):
if i % 3 == 0:
final = string.replace(string[i], "")
print(final)
I was asked the question: "Given a string, delete all its characters whose indices are divisible by 3."
The answer for the input Python is yton. However, my code gives Pyton.
The code makes sense to me but I'm a beginner. Any help?
The problem is that while you are looping, you are overriding the final variable every time the index is divisible by 3.
Instead, try defining the final variable before you start the loop, and add the letters as you loop over them, and only when they their index is NOT divisible by 3 (thus, ignoring the ones where the index IS divisible by 3).
Something like this should work:
string = input()
final = ""
for i in range(len(string)):
if i % 3 != 0:
final += string[i]
print(final)
In your current code, final is used through each iteration of the loop. It continues updating by replacing one character. In each iteration, final is replaced by a different string with one letter from string removed. After the loop has completed, it effectively only replaced one letter, which in this case is "h".
Use this instead (thanks to Mateen Ulhaq for the idea):
print("".join(x for i, x in enumerate(input()) if i % 3 != 0))
string=input()
final=""
for i in range(len(string)):
if i % 3 != 0:
final+=string[i]
print(final)
In your code, the line final = string.replace(string[i], "") would run like this.
Supposing the input is "hellobaby":
i=0, final="ellobaby"
i=3, final="helobaby"
i=6, final="hellobby"
I have to enter a string, remove all spaces and print the string without vowels. I also have to print a string of all the removed vowels.
I have gotten very close to this goal, but for some reason when I try to remove all the vowels it will not remove two vowels in a row. Why is this? Please give answers for this specific block of code, as solutions have helped me solve the challenge but not my specific problem
# first define our function
def disemvowel(words):
# separate the sentence into separate letters in a list
no_v = list(words.lower().replace(" ", ""))
print no_v
# create an empty list for all vowels
v = []
# assign the number 0 to a
a = 0
for l in no_v:
# if a letter in the list is a vowel:
if l == "a" or l == "e" or l == "i" or l == "o" or l == "u":
# add it to the vowel list
v.append(l)
#print v
# delete it from the original list with a
del no_v[a]
print no_v
# increment a by 1, in order to keep a's position in the list moving
else:
a += 1
# print both lists with all spaces removed, joined together
print "".join(no_v)
print "".join(v)
disemvowel(raw_input(""))
Mistakes
So there are a lot of other, and perhaps better approaches to solve this problem. But as you mentioned I just discuss your failures or what you can do better.
1. Make a list of input word
There are a lot of thins you could do better
no_v = list(words.lower().replace(" ", ""))
You don't replaces all spaces cause of " " -> " " so just use this instead
no_v = list(words.lower().translate( None, string.whitespace))
2. Replace for loop with while loop
Because if you delete an element of the list the for l in no_v: will go to the next position. But because of the deletion you need the same position, to remove all the vowels in no_v and put them in v.
while a < len(no_v):
l = no_v[a]
3. Return the values
Cause it's a function don't print the values just return them. In this case replace the print no_v print v and just return and print them.
return (no_v,v) # returning both lists as tuple
4. Not a mistake but be prepared for python 3.x
Just try to use always print("Have a nice day") instead of print "Have a nice day"
Your Algorithm without the mistakes
Your algorithm now looks like this
import string
def disemvowel(words):
no_v = list(words.lower().translate( None, string.whitespace))
v = []
a = 0
while a < len(no_v):
l = no_v[a]
if l == "a" or l == "e" or l == "i" or l == "o" or l == "u":
v.append(l)
del no_v[a]
else:
a += 1
return ("".join(no_v),"".join(v))
print(disemvowel("Stackoverflow is cool !"))
Output
For the sentence Stackoverflow is cool !\n it outputs
('stckvrflwscl!', 'aoeoioo')
How I would do this in python
Not asked but I give you a solution I would probably use. Cause it has something to do with string replacement, or matching I would just use regex.
def myDisemvowel(words):
words = words.lower().translate( None, string.whitespace)
nv = re.sub("[aeiou]*","", words)
v = re.sub("[^a^e^i^o^u]*","", words)
return (nv, v)
print(myDisemvowel("Stackoverflow is cool !\n"))
I use just a regular expression and for the nv string I just replace all voewls with and empty string. For the vowel string I just replace the group of all non vowels with an empty string. If you write this compact, you could solve this with 2 lines of code (Just returning the replacement)
Output
For the sentence Stackoverflow is cool !\n it outputs
('stckvrflwscl!', 'aoeoioo')
You are modifying no_v while iterating through it. It'd be a lot simpler just to make two new lists, one with vowels and one without.
Another option is to convert it to a while loop:
while a < len(no_v):
l = no_v[a]
This way you have just a single variable tracking your place in no_v instead of the two you currently have.
For educational purposes, this all can be made significantly less cumbersome.
def devowel(input_str, vowels="aeiou"):
filtered_chars = [char for char in input_str
if char.lower() not in vowels and not char.isspace()]
return ''.join(filtered_chars)
assert devowel('big BOOM') == 'bgBM'
To help you learn, do the following:
Define a function that returns True if a particular character has to be removed.
Using that function, loop through the characters of the input string and only leave eligible characters.
In the above, avoid using indexes and len(), instead iterate over characters, as in for char in input_str:.
Learn about list comprehensions.
(Bonus points:) Read about the filter function.
I have been learning python through code academy. It asked me to create a if else statement that prints the users input, but if there is no input print "empty". I did pass the tutorial but when there is a user input it prints both the users input and "empty".
here is my code:
print "Welcome to the English to Pig Latin translator!"
original = raw_input("what is your name?")
length = len("original")
if length > 0:
print original
else: length < 0
print "empty"
Notice that print under else is not indented. I thought you had to indented it, but when i do it gives me an error.
You seem to have a couple issues with your code. First I believe you should change your assignment of length to:
length = len(original)
That way you get the correct length of the string you binded to the name original. If you use len("original"), you will pass the string "original" to the len() function, and it will just count the characters in that string. Regardless of what the user of your program inputted, your length will always be 8, since that's how many characters are in the string "original"
Also keep in mind that the len() function will never return a negative value. So testing if length < 0 doesn't make sense. You probably wanted to test if it equals 0. It seems as if you were trying to use an elif statement.
Try:
print "Welcome to the English to Pig Latin translator!"
original = raw_input("what is your name?")
length = len(original)
if length > 0:
print original
elif length == 0:
print "empty"
elif statements let you test conditions just like if statements. elif is short for "else if".
Furthermore, as #chepner pointed out in the comments, there is no need to even test the second condition at all. You can just as easily go with:
else:
print "empty"
The else syntax is not followed by any condition. It automatically enters the indented block of code if the other conditions evaluate to False.
Was the length < 0 intended to be a comment? You need the comment character #.
else: # length < 0
print "empty"
It's wrong anyway, it should be length <= 0.
Without a comment character it was being interpreted as the statement to use as the else clause. It didn't create an error since length < 0 just generates a boolean result that you didn't do anything with, but it prevented you from having another statement as part of the else.
else: length < 0
print "empty"
should be
elif length == 0:
print "empty"
Python has significant whitespace, things that are indented the same are in the same scope.
First off, it is not len("original") it is len(original). If you use quotes you are making it a constant value, "original", rather than a variable named original.
Second, instead of checking the length of the string you should use this
if original:
# Type here what happens if the string is not empty
else:
# Here what happens if the string is empty
By Python standard any collection, including strings, equals false if it is empty(aka contains no elements) and true if it contains any elements.
There is a a statement after else.
else: length < 0
print "empty"
Maybe you are looking for an (elif is another way to check another condition after the if if the previous if fails.)
elif length <= 0:
or just a plain
else:
print "empty"
it will never go past zero anyways you could have a == conditional for zero and it would work.
elif length == 0:
this is probably the best way there is no need to check another condition.
if length > 0:
print original
else:
print "empty"
also just a side note length = len("original")
there is not suppose to be quotation marks around original because its a variable :). You will just be getting the string "original" not the actual stuff inside of the variable.
The end result ..
print "Welcome to the English to Pig Latin translator!"
original = raw_input("what is your name?")
length = len(original)
if length > 0:
print original
else:
print "empty"
To check if there is an item in a list, simply do this:
if List_name:
{code here}
...
So, your code should very simply look like this:
print "Welcome to the English to Pig Latin translator!"
original = raw_input("what is your name?")
if original:
print original
else:
print "Empty"
It's that easy :D
So i had to write a program that asks for a user input (which should be a 3 letter string) and it outputs the six permutations of the variations of the placements of the letters inside the string. However, my professor wants the output to be surrounded by curly brackets whereas mine is a list (so it is square brackets). How do i fix this? Also, how do I check if none of the letters in the input repeat so that the main program keeps asking the user to enter input and check it for error?
Thank you
The only datatype im aware of that 'natively' outputs with { } is a dictionary, which doesnt seem to apply here. I would just write a small function to output your lists in the desired fashion
>>> def curlyBracketOutput(l):
x = ''
for i in l: x += i
return '{' + x + '}'
>>> curlyBracketOutput(['a','b','c'])
'{abc}'
ok, for one thing, as everyone here has said, print '{'. other than that, you can use the following code in your script to check for repeated words,
letterlist = []
def takeInput(string):
for x in string:
if x not in letterlist:
letterlist.append(x)
else:
return 0
return 1
then as for your asking for input and checking for errors, you can do that by,
while(True): #or any other condition
string = input("Enter 3 letter string")
if len(string)!=3:
print("String size inadequate")
continue
if takeInput(string):
arraylist = permutation(string) #--call permutation method here
#then iterate the permutations and print them in {}
for x in arraylist: print("{" + x + "}")
else:
print("At least one of the letters already used")
The answer to both question is to use a loop.
Print the "{" and then loop through all the elements printing them.
But the input inside a loop and keep looping until you get what you want.
Curly brackets refers to a dict?
I think a
list(set(the_input))
should give you a list of unique letters. to check if they occur more than once
and
theinput.count(one_letter) > 1
should tell you if there is mor than one.
>>> chars = ['a','b','c']
>>> def Output(chars):
... return "{%s}" % ''.join(chars)
...
>>> print Output(chars)
{abc}
>>>
Or just do something tremendously kludgy:
print repr(YourExistingOutput).replace("[", "{").replace("]", "}")