I want to add a point before each letter.
Unfortunately, the point is after each letter. How to insert the point before each letter?
myText = str(input())
vowels = ['a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U']
result = ''
for letter in myText:
if letter not in vowels:
result = result + letter
for i in result:
result = result + '.'
break
print(result.lower())
A shorter and faster solution relying on the power of regular expressions:
import re
my_text = "sample"
re.sub(r"[aeiou]*([^aeiou])[aeiou]*", r".\1", my_text.lower())
This reads: “delete vowels, and prefix each remaining letter with a dot”.
Removing all vowels and append "." before each letter.
myText = "sample"
vowels = ['a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U']
result = ''
for letter in myText:
if letter not in vowels:
result = result + '.' + letter
print(result.lower())
# .s.m.p.l
Fian's answer is probably the best one. Here's my attempt:
text = 'audfijsdfmsomlasn'
vowels = {'a', 'e', 'i', 'o', 'u', 'A', 'E', 'I', 'O', 'U'}
result = ''
for letter in text:
if letter in vowels:
result += '.'
result += letter
print(result)
Yet another way:
import re
myText = "sample"
result = '.' + '.'.join(list(re.sub('[aeiou]', "", myText, flags=re.I)))
print(result)
Explanation:
re.sub removes the not required letters (case insensitive = re.I)
list makes the string an array of characters
join places a dot in between
'.' + adds the missing dot in front of the first character
string = input()
vowels = ['a', 'A', 'i', 'I', 'o', 'O', 'u', 'U', 'e', 'E']
translation = string.maketrans({i: '' for i in vowels})
string = string.translate(translation)
result = ''
for letter in string:
result = result + '.' + letter
print(result.lower())
vowels = "aeiou"
consonants = "bcdfghjklmnpqrstvwxyz"
I'm trying to make a function that would only return constants followed by vowels in a list.
So for example:
f("therapist")
>>>["he", "ra", "pi"]
f("apple")
>>>["le"]
So it's only when a vowel follows a consonant and it returns both the consonant and the vowel in a list.
I was thinking it would be something along the lines of:
def f(word):
for consonant in word:
for vowel in word:
But I don't know how to work with the order and test whether the vowel is after the consonant. Thanks in advance
You can use enumerate with a starting index of 1, checking if the current ele is a consonant and the next chaarcter word[i] is a vowel.
def f(word):
vowels = {"a", "e", "i", "o", "u"}
consonants = {'t', 'b', 'm', 'h', 'y', 'w', 'z', 'p', 'v', 'd', 'g', 'k', 'j', 'n', 'r', 'q', 'x', 'c', 's','f', 'l'}
return [ele + word[i] for i, ele in enumerate(word[:-1], 1)
if word[i] in vowels and ele in consonants ]
Or using loops keep track of the last character and compare:
def f(word):
vowels = {"a", "e", "i", "o", "u"}
consonants = {'t', 'b', 'm', 'h', 'y', 'w', 'z', 'p', 'v', 'd', 'g', 'k', 'j', 'n', 'r', 'q', 'x', 'c', 's','f', 'l'}
pairs = []
it = iter(word)
# get first char
prev = next(it,"")
# iterate over the rest starting at the second char
for ch in it:
# compare prev and current
if prev in consonants and ch in vowels:
pairs.append(prev + ch)
# update prev
prev = ch
return pairs
You can use regex :
>>> import re
>>> def f(s):
... return re.findall(r'[bcdfghjklmnpqrstvwxyz][aeiou]',s)
...
>>> f('therapist')
['he', 'ra', 'pi']
Also you can use zip within a list comprehension :
>>> def f(s):
... return [''.join((i,j)) for i,j in zip(s,s[1:]) if i in 'bcdfghjklmnpqrstvwxyz' and j in 'aeiou']
...
>>> s='therapist'
>>> f(s)
['he', 'ra', 'pi']
Well, you could use a regular expression for this.
I think the real question is: "Does 'y' constitute as a vowel?"
import re
def f(word):
return re.findall('[bcdfghjklmnpqrstvwxyz][aeiou]', word)
print(f("therapist")) # Prints ['he', 'ra', 'pi']
Try this using re
import re
def f(word):
return re.findall(r"(?![aeiou$])\w[aeiou]",word)
>>>f('therapist')
['he', 'ra', 'pi']
This question already has answers here:
Strange result when removing item from a list while iterating over it
(8 answers)
Closed 7 years ago.
in this code I am trying to create a function anti_vowel that will remove all vowels (aeiouAEIOU) from a string. I think it should work ok, but when I run it, the sample text "Hey look Words!" is returned as "Hy lk Words!". It "forgets" to remove the last 'o'. How can this be?
text = "Hey look Words!"
def anti_vowel(text):
textlist = list(text)
for char in textlist:
if char.lower() in 'aeiou':
textlist.remove(char)
return "".join(textlist)
print anti_vowel(text)
You're modifying the list you're iterating over, which is bound to result in some unintuitive behavior. Instead, make a copy of the list so you don't remove elements from what you're iterating through.
for char in textlist[:]: #shallow copy of the list
# etc
To clarify the behavior you're seeing, check this out. Put print char, textlist at the beginning of your (original) loop. You'd expect, perhaps, that this would print out your string vertically, alongside the list, but what you'll actually get is this:
H ['H', 'e', 'y', ' ', 'l', 'o', 'o', 'k', ' ', 'W', 'o', 'r', 'd', 's', '!']
e ['H', 'e', 'y', ' ', 'l', 'o', 'o', 'k', ' ', 'W', 'o', 'r', 'd', 's', '!']
['H', 'y', ' ', 'l', 'o', 'o', 'k', ' ', 'W', 'o', 'r', 'd', 's', '!'] # !
l ['H', 'y', ' ', 'l', 'o', 'o', 'k', ' ', 'W', 'o', 'r', 'd', 's', '!']
o ['H', 'y', ' ', 'l', 'o', 'o', 'k', ' ', 'W', 'o', 'r', 'd', 's', '!']
k ['H', 'y', ' ', 'l', 'o', 'k', ' ', 'W', 'o', 'r', 'd', 's', '!'] # Problem!!
['H', 'y', ' ', 'l', 'o', 'k', ' ', 'W', 'o', 'r', 'd', 's', '!']
W ['H', 'y', ' ', 'l', 'o', 'k', ' ', 'W', 'o', 'r', 'd', 's', '!']
o ['H', 'y', ' ', 'l', 'o', 'k', ' ', 'W', 'o', 'r', 'd', 's', '!']
d ['H', 'y', ' ', 'l', 'k', ' ', 'W', 'o', 'r', 'd', 's', '!']
s ['H', 'y', ' ', 'l', 'k', ' ', 'W', 'o', 'r', 'd', 's', '!']
! ['H', 'y', ' ', 'l', 'k', ' ', 'W', 'o', 'r', 'd', 's', '!']
Hy lk Words!
So what's going on? The nice for x in y loop in Python is really just syntactic sugar: it still accesses list elements by index. So when you remove elements from the list while iterating over it, you start skipping values (as you can see above). As a result, you never see the second o in "look"; you skip over it because the index has advanced "past" it when you deleted the previous element. Then, when you get to the o in "Words", you go to remove the first occurrence of 'o', which is the one you skipped before.
As others have mentioned, list comprehensions are probably an even better (cleaner, clearer) way to do this. Make use of the fact that Python strings are iterable:
def remove_vowels(text): # function names should start with verbs! :)
return ''.join(ch for ch in text if ch.lower() not in 'aeiou')
Other answers tell you why for skips items as you alter the list. This answer tells you how you should remove characters in a string without an explicit loop, instead.
Use str.translate():
vowels = 'aeiou'
vowels += vowels.upper()
text.translate(None, vowels)
This deletes all characters listed in the second argument.
Demo:
>>> text = "Hey look Words!"
>>> vowels = 'aeiou'
>>> vowels += vowels.upper()
>>> text.translate(None, vowels)
'Hy lk Wrds!'
>>> text = 'The Quick Brown Fox Jumps Over The Lazy Fox'
>>> text.translate(None, vowels)
'Th Qck Brwn Fx Jmps vr Th Lzy Fx'
In Python 3, the str.translate() method (Python 2: unicode.translate()) differs in that it doesn't take a deletechars parameter; the first argument is a dictionary mapping Unicode ordinals (integer values) to new values instead. Use None for any character that needs to be deleted:
# Python 3 code
vowels = 'aeiou'
vowels += vowels.upper()
vowels_table = dict.fromkeys(map(ord, vowels))
text.translate(vowels_table)
You can also use the str.maketrans() static method to produce that mapping:
vowels = 'aeiou'
vowels += vowels.upper()
text.translate(text.maketrans('', '', vowels))
Quoting from the docs:
Note: There is a subtlety when the sequence is being modified by the
loop (this can only occur for mutable sequences, i.e. lists). An
internal counter is used to keep track of which item is used next, and
this is incremented on each iteration. When this counter has reached
the length of the sequence the loop terminates. This means that if the
suite deletes the current (or a previous) item from the sequence, the
next item will be skipped (since it gets the index of the current item
which has already been treated). Likewise, if the suite inserts an
item in the sequence before the current item, the current item will be
treated again the next time through the loop. This can lead to nasty
bugs that can be avoided by making a temporary copy using a slice of
the whole sequence, e.g.,
for x in a[:]:
if x < 0: a.remove(x)
Iterate over a shallow copy of the list using [:]. You're modifying a list while iterating over it, this will result in some letters being missed.
The for loop keeps track of index, so when you remove an item at index i, the next item at i+1th position shifts to the current index(i) and hence in the next iteration you'll actually pick the i+2th item.
Lets take an easy example:
>>> text = "whoops"
>>> textlist = list(text)
>>> textlist
['w', 'h', 'o', 'o', 'p', 's']
for char in textlist:
if char.lower() in 'aeiou':
textlist.remove(char)
Iteration 1 : Index = 0.
char = 'W' as it is at index 0. As it doesn't satisfies that condition you'll do noting.
Iteration 2 : Index = 1.
char = 'h' as it is at index 1. Nothing more to do here.
Iteration 3 : Index = 2.
char = 'o' as it is at index 2. As this item satisfies the condition so it'll be removed from the list and all the items to it's right will shift one place to the left to fill the gap.
now textlist becomes :
0 1 2 3 4
`['w', 'h', 'o', 'p', 's']`
As you can see the other 'o' moved to index 2, i.e the current index so it'll be skipped in the next iteration. So, this is the reason some items are bring skipped in your iteration. Whenever you remove an item the next item is skipped from the iteration.
Iteration 4 : Index = 3.
char = 'p' as it is at index 3.
....
Fix:
Iterate over a shallow copy of the list to fix this issue:
for char in textlist[:]: #note the [:]
if char.lower() in 'aeiou':
textlist.remove(char)
Other alternatives:
List comprehension:
A one-liner using str.join and a list comprehension:
vowels = 'aeiou'
text = "Hey look Words!"
return "".join([char for char in text if char.lower() not in vowels])
regex:
>>> import re
>>> text = "Hey look Words!"
>>> re.sub('[aeiou]', '', text, flags=re.I)
'Hy lk Wrds!'
You're modifying the data you're iterating over. Don't do that.
''.join(x for x in textlist in x not in VOWELS)
text = "Hey look Words!"
print filter(lambda x: x not in "AaEeIiOoUu", text)
Output
Hy lk Wrds!
You're iterating over a list and deleting elements from it at the same time.
First, I need to make sure you clearly understand the role of char in for char in textlist: .... Take the situation where we have reached the letter 'l'. The situation is not like this:
['H', 'e', 'y', ' ', 'l', 'o', 'o', 'k', ' ', 'W', 'o', 'r', 'd', 's', '!']
^
char
There is no link between char and the position of the letter 'l' in the list. If you modify char, the list will not be modified. The situation is more like this:
['H', 'e', 'y', ' ', 'l', 'o', 'o', 'k', ' ', 'W', 'o', 'r', 'd', 's', '!']
^
char = 'l'
Notice that I've kept the ^ symbol. This is the hidden pointer that the code managing the for char in textlist: ... loop uses to keep track of its position in the loop. Every time you enter the body of the loop, the pointer is advanced, and the letter referenced by the pointer is copied into char.
Your problem occurs when you have two vowels in succession. I'll show you what happens from the point where you reach 'l'. Notice that I've also changed the word "look" to "leap", to make it clearer what's going on:
advance pointer to next character ('l') and copy to char
['H', 'e', 'y', ' ', 'l', 'e', 'a', 'p', ' ', 'W', 'o', 'r', 'd', 's', '!']
-> ^
char = 'l'
char ('l') is not a vowel, so do nothing
advance pointer to next character ('e') and copy to char
['H', 'e', 'y', ' ', 'l', 'e', 'a', 'p', ' ', 'W', 'o', 'r', 'd', 's', '!']
-> ^
char = 'e'
char ('e') is a vowel, so delete the first occurrence of char ('e')
['H', 'e', 'y', ' ', 'l', 'e', 'a', 'p', ' ', 'W', 'o', 'r', 'd', 's', '!']
^
['H', 'e', 'y', ' ', 'l', 'a', 'p', ' ', 'W', 'o', 'r', 'd', 's', '!']
^
['H', 'e', 'y', ' ', 'l', <- 'a', 'p', ' ', 'W', 'o', 'r', 'd', 's', '!']
^
['H', 'e', 'y', ' ', 'l', 'a', 'p', ' ', 'W', 'o', 'r', 'd', 's', '!']
^
advance pointer to next character ('p') and copy to char
['H', 'e', 'y', ' ', 'l', 'a', 'p', ' ', 'W', 'o', 'r', 'd', 's', '!']
-> ^
char = 'p'
When you removed the 'e' all the characters after the 'e' moved one place to the left, so it was as if remove had advanced the pointer. The result is that you skipped past the 'a'.
In general, you should avoid modifying lists while iterating over them. It's better to construct a new list from scratch, and Python's list comprehensions are the perfect tool for doing this. E.g.
print ''.join([char for char in "Hey look Words" if char.lower() not in "aeiou"])
But if you haven't learnt about comprehensions yet, the best way is probably:
text = "Hey look Words!"
def anti_vowel(text):
textlist = list(text)
new_textlist = []
for char in textlist:
if char.lower() not in 'aeiou':
new_textlist.append(char)
return "".join(new_textlist)
print anti_vowel(text)
List Comprehensions:
vowels = 'aeiou'
text = 'Hey look Words!'
result = [char for char in text if char not in vowels]
print ''.join(result)
Others have already explained the issue with your code. For your task, a generator expression is easier and less error prone.
>>> text = "Hey look Words!"
>>> ''.join(c for c in text if c.lower() not in 'aeiou')
'Hy lk Wrds!'
or
>>> ''.join(c for c in text if c not in 'AaEeIiOoUu')
'Hy lk Wrds!'
however, str.translate is the best way to go.
You shouldn't delete items from list you iterating through:
But you can make new list from the old one with list comprehension syntax. List comprehension is very useful in this situation. You can read about list comprehension here
So you solution will look like this:
text = "Hey look Words!"
def anti_vowel(text):
return "".join([char for char in list(text) if char.lower() not in 'aeiou'])
print anti_vowel(text)
It's pretty, isn't it :P
Try to not use the list() function on a string. It will make things a lot more complicated.
Unlike Java, in Python, strings are considered as arrays. Then, try to use an index for loop and del keyword.
for x in range(len(string)):
if string[x].lower() in "aeiou":
del string[x]
This is my current code:
key = input("Enter the key: ")
sent = input("Enter a sentence: ")
print()# for turnin
print()
print("With a key of:",key)
print("Original sentence:",sent)
print()
#split = sent.split()
blank = [ ]
for word in sent:
for ch in word:
blank = blank + ch.split()
print(blank)
print()
What i have now gives me a list of all the letters in my sentence, but no spaces. If i use this...
for word in sent:
for ch in word:
print(ch.split())
It gives me a list of all characters including the spaces. Is there to get this result and have it equal a variable?
If you just want a list of all characters in the sentence, use
chars = list(sent)
What you're doing is definitely not what you think you're doing.
for word in sent:
This doesn't loop over the words. This loops over the characters. This:
for word in sent.split()
would loop over the words.
for ch in word:
Since word is a character already, this loops over a single character. If it weren't for the fact that characters are represented as length-1 strings, this would throw some kind of error.
sent is of type string. and when you iterate over a string this way:
for word in sent:
you get the individual characters, not the words.
Then you iterate over a single char:
for ch in word:
and get that very same char (!).
And then with that split() call you convert a non-blank character, say 'x' into a list with itself as element (['x']) and a blank characters into the empty list.
You probably want something along the lines of:
for word in sent.split():
....
But if what you want is to build a list of words, no need to iterate, that's exactly what sent.split() will get you!
And if what you want is a list of chars, do list(sent).
From help(str.split):
split(...)
S.split(sep=None, maxsplit=-1) -> list of strings
Return a list of the words in S, using sep as the
delimiter string. If maxsplit is given, at most maxsplit
splits are done. If sep is not specified or is None, any
whitespace string is a separator and empty strings are
removed from the result.
If you want individual characters of a string, pass it to list.
>>> list('This is a string.')
['T', 'h', 'i', 's', ' ', 'i', 's', ' ', 'a', ' ', 's', 't', 'r', 'i', 'n', 'g', '.']
I'm not 100% sure what you're asking, but it seems like....
blank = [ch for ch in sent]
...that's all you need....
Let me give you some sample Ins and Outs and see if that's what you want.
IN = "Hello world!"
OUT =>
['H', 'e', 'l', 'l', 'o', ' ', 'w', 'o', 'r', 'l', 'd', '!']
Is that right?
string = "This here is a string"
>>> x = list(string) # I think list() is what you are looking for... It's not clear
>>> print x
['T', 'h', 'i', 's', ' ', 'h', 'e', 'r', 'e', ' ', 'i', 's', ' ', 'a', ' ', 's', 't', 'r', 'i', 'n', 'g']
>>> print string.split() # default arg is a space
['This', 'here', 'is', 'a', 'string']