I'm trying to get a substring in a for loop. For that I'm using this:
for peoject in subjects:
peoject_name = peoject.content
print(peoject_name, " : ", len(peoject_name), " : ", len(peoject_name.split('-')[1]))
I have some projects that don't have any "-" in the sentence. How can I deal with this?
I'm getting this issue:
builtins.IndexError: list index out of range
for peoject in subjects:
try:
peoject_name = peoject.content
print(peoject_name, " : ", len(peoject_name), " : ", len(peoject_name.split('-')[1]))
except IndexError:
print("this line doesn't have a -")
you could just check if there is a '-' in peoject_name:
for peoject in subjects:
peoject_name = peoject.content
if '-' in peoject_name:
print(peoject_name, " : ", len(peoject_name), " : ",
len(peoject_name.split('-')[1]))
else:
# something else
You have a few options, depending on what you want to do in the case where there is no hyphen.
Either select the last item from the split via [-1], or use a ternary statement to apply alternative logic.
x = 'hello-test'
print(x.split('-')[1]) # test
print(x.split('-')[-1]) # test
y = 'hello'
print(y.split('-')[-1]) # hello
print(y.split('-')[1] if len(y.split('-'))>=2 else y) # hello
print(y.split('-')[1] if len(y.split('-'))>=2 else '') # [empty string]
Related
I need to move a whitespace in a string one position to the right.
This is my code:
for i in range(0,len(resultaat)):
if resultaat[i] == " ":
string = resultaat[:i] + resultaat[i+1] + " " + resultaat[i+2:]
E.g.:
If resultaat =
"TH EZE NO FPYTHON."
Than my output needs to be:
'THE ZEN OF PYTHON.'
, but the output that I get is:
"TH EZE NO F PYTHON."
I think this happened because the loop undid the action where it moved the previous space.
I don't know how to fix this problem.
Can someone help me with this?
Thanks!
Each time through the loop you're getting slices of the original resultaat string, without the changes you've made for previous iterations.
You should copy resultaat to string first, then use that as the source of each slice so you accumulate all the changes.
string = resultaat
for i in range(0,len(resultaat)):
if resultaat[i] == " ":
string = string[:i] + string[i+1] + " " + string[i+2:]
You could do something like this:
# first get the indexes that the character you want to merge
indexes = [i for i, c in enumerate(resultaat) if c == ' ']
for i in indexes: # go through those indexes and swap the characters as you have done
resultaat = resultaat[:i] + resultaat[i+1] + " " + resultaat[i+2:] # updating resultaat each time you want to swap characters
Assuming the stated input value actually has one more space than is actually needed then:
TXT = "TH EZE NO FPYTHON."
def process(s):
t = list(s)
for i, c in enumerate(t[:-1]):
if c == ' ':
t[i+1], t[i] = ' ', t[i+1]
return ''.join(t)
print(process(TXT))
Output:
THE ZEN OF PYTHON.
How do I print a specific character from a string in Python? I am still learning and now trying to make a hangman like program. The idea is that the user enters one character, and if it is in the word, the word will be printed with all the undiscovered letters as "-".
I am not asking for a way to make my idea/code of the whole project better, just a way to, as i said, print that one specific character of the string.
print(yourstring[characterposition])
Example
print("foobar"[3])
prints the letter b
EDIT:
mystring = "hello world"
lookingfor = "l"
for c in range(0, len(mystring)):
if mystring[c] == lookingfor:
print(str(c) + " " + mystring[c]);
Outputs:
2 l
3 l
9 l
And more along the lines of hangman:
mystring = "hello world"
lookingfor = "l"
for c in range(0, len(mystring)):
if mystring[c] == lookingfor:
print(mystring[c], end="")
elif mystring[c] == " ":
print(" ", end="")
else:
print("-", end="")
produces
--ll- ---l-
all you need to do is add brackets with the char number to the end of the name of the string you want to print, i.e.
text="hello"
print(text[0])
print(text[2])
print(text[1])
returns:
h
l
e
Well if you know the character you want to search you can use this approach.
i = character looking for
input1 = string
if i in input1:
print(i)
you can change the print statement according to your logic.
name = "premier league"
for x in name:
print(x)
Result shown below:-
To print specific characters of the given string. For example to print 'l' from the given string
name = "premier league"
for x in name:
if x == "l":
print("element found: "+x)
I am having issues with formatting and understand lists and strings when it comes to classes.
So I have this code here:
class User:
def __init__(self,title):
self.tile=tile
self.rank={}
def addCard(self,compID,number):
if compID in self.cards and number > self.cards[compID]:
self.cards[compID]=number
elif compID not in self.cards:
self.cards[compID]=number
def __str__(self):
self.cardList = []
for compID, number in self.cards.items():
final = compID + "-" + str(number)
self.cardList.append(temp)
self.cardList.sort()
return self.tile + ":" + " " + "Card scores:" + str(self.cardList)
so my result looks like this:
OUTPUT 1:
Cpt.Fred: Card scores: ['diamond-22', 'hearts-4', 'spades-3']
Lt.Connor: Card scores: ['diamond-43']
I am trying to make my result look like this:
OUTPUT 2:
Cpt.Fred: Card scores: [ diamond-22, hearts-4, spades-3 ]
Lt.Connor: Card scores: [ diamond-43 ]
The data is not whats important, its how to get rid of the " ' " at the beginning and the end of the results. I think it has something to do with my last def() statement but I have been trying to format it every way with no luck. Can anyone help turn the first output to look like the second output?
Instead of calling str(self.cardList), you should do:
return "%s: Card scores: [%s]" % (self.title, ", ".join(self.cardList))
The problem is that str on a list calls repr (includes quotes) on the list's elements, whereas you just want the str of the elements joined by commas.
You may need to process and print the list manually. Here's one way you could do that.
def printList(cardlist):
printstr = "[" # Start with [.
for card in cardlist:
printstr += card + ", " "card, "
printstr = printstr[:-2] + "]" # Remove the last ", " before adding the last ].
The problem is in "str(self.cardList)". It prints out the list as is, meaning it puts strings into quotations to distinguished them from numbers and other objects.
def list_to_string(some_list):
str = "["
for i in range(len(some_list)):
str += some_list[i]
if i < len(some_list)-1:
str += ", "
str += "]"
return str
this would parse your list into a string without the quotations
I've been working on HTTLCS and am having some difficulty finishing up the problem.
Solving a problem was not much of an issue, but I have trouble returning my result as a string rather than the tuple data type.
Here is my code:
def wordCount(paragraph):
splited = paragraph.split()
wordnum = len(splited)
eWord = []
for aWord in splited:
if "e" in aWord:
eWord.append(aWord)
eWordnum = len(eWord)
percent = round(eWordnum / wordnum * 100,2)
return "Your text contains", wordnum, "words, of which" , eWordnum , "(" , percent , "%)" , "contains an 'e'."
print(wordCount(p))
Python outputs ('Your text contains', 108, 'words, of which', 50, '(', 46.3, '%)', "contains an 'e'.") which is a tuple, not a string.
I know I can just put print at the end of the function and call the function without print() statement, but how do I solve this with a return statement?
It's because you're using commas in your return statement, which Python is interpreting as a tuple. Try using format() instead:
def wordCount(paragraph):
splited = paragraph.split()
wordnum = len(splited)
eWord = []
for aWord in splited:
if "e" in aWord:
eWord.append(aWord)
eWordnum = len(eWord)
percent = round(eWordnum / wordnum * 100,2)
return "Your text contains {0} words, of which {1} ({2}%) contains an 'e'".format(wordnum, eWordnum, percent)
>>> wordCount("doodle bugs")
"Your text contains 2 words, of which 1 (0.0%) contains an 'e'"
return "Your text contains {0} words, of which {1} ({2}%) contains an 'e'.".format(wordnum,eWordnum,percent)
return "Your text contains " + str(wordnum) +
" words, of which " + str(eWordnum) +
" (" + str(percent) + "%)" + " contains an 'e'."
or
return "Your text contains %s words, of which %s (%s%%) contains an 'e'."
% (wordnum, eWordnum, percent)
In the first case, you do a string concatenation and you have to convert wordnum, eWordnum and other variables that are numeric ones, into str (by doing str(variableName)) to allow the concatenation (and for haven't runtime error)
In the second case, you do a string replacement that means that you give some kind of "placeholder" %s (that means string) and you replace them with tuple argument that follows the % symbol
If you return something separate by , you'll return a tuple (as you can see)
return "Your text contains %s words, of which %s (%s%%) contains an 'e'." % (wordnum, eWordnum, percent)
A for loop might work, though you would have to format the strings to add spaces to them.
for item in tuplename: print item,
Make sure to keep the comma after item, because that prints it on the same line.
def wordCount(paragraph):
splited = paragraph.split()
wordnum = len(splited)
eWord = []
for aWord in splited:
if "e" in aWord:
eWord.append(aWord)
eWordnum = len(eWord)
percent = round(eWordnum / wordnum * 100,2)
dummy = "Your text contains {0} words, of which {1} {2} contains an 'e'.".format(wordnum,eWordnum, percent)
return dummy
print(wordCount(p))
try this :
return "Your text contains %(wordnum)s words, of which %(ewordnum)s (%(percent)s %%), contains an 'e'."%locals()
using %(variable_name)s as string formatting is often easier to maintain.
how about this
return "Your text contains " + wordnum + " words, of which " + eWordnum + " (" + percent + "%) " + " contains an 'e'."
replace the commas with "+", this should work.
When I run the code below I get:
Thank you for joining, ['cars', 'gas', 'jewelry']but['bus', 'join'] are not keywords.
How can I effectively turn the lists in to just strings to be printed? I suspect I may need a regular expression... this time :)
import re
pattern = re.compile('[a-z]+', re.IGNORECASE)
text = "join cars jewelry gas bus"
keywordset = set(('cars', 'jewelry', 'gas', 'food', 'van', 'party', 'shoes'))
words = pattern.findall(text.lower())
notkeywords = list(set(words) - keywordset)
keywords = list(keywordset & set(words))
if notkeywords == ['join']:
print "Thank you for joining keywords " + str(keywords) + "!"
else:
print "Thank you for joining, " + str(keywords) + "but" + str(notkeywords) + " are not keywords."
To convert list to strings use str.join like this
print "Thank you for joining keywords " + ",".join(keywords) + "!"
This if notkeywords == ['join']: is not a way to compare list elements.
>>> mylist = [1,2]
>>> mylist == 1
False
you should in operator to check for equality.
>>> mylist = [1,2]
>>> 1 in mylist
True
Just use someString.join(list):
if notkeywords == ['join']:
print "Thank you for joining keywords " + ", ".join(keywords) + "!"
else:
print "Thank you for joining, " + ", ".join(keywords) + "but" + ", ".join(notkeywords) + " are not keywords."
If I understood your question correctly, you'll want to use the .join() string method to combine the list before printing it.
For example:
', '.join(my_list)
will give you comma separated output. ', ' can be whatever kind of separator you like.