I am learning how to program in Python 3 and I am working on a project that lets you buy a ticket to a movie. After that you can see your shopping cart with all the tickets that you have bought.
Now, I want after each printed line to add a integer.
For example: 1. Movie1 , 2. Movie2 , etc..
Here is my code that I use to print the films:
if choice == 3:
#try:
print("Daca doresti sa vezi ce filme sunt valabile, scrie exit.")
bilet = str(input("Ce film doresti sa vizionezi?: ").title())
pret = films[bilet]["price"]
cumperi = input("Doresti sa adaugi in cosul de cumparaturi {}$ (y/n)?".format(bilet)).strip().lower()
if cumperi == "y":
bani[0] -= pret
cos.append(bilet)
if choice == 4:
print (*cos, sep="\n")
You can use an integral variable and increase it's value whenever you perform a task.
example set count = 0 and when you does a task place this there count += 1.
Related
I am trying to create an anime series search using the tool anilistpython, but I am not able to ignore the newline character in the plot tag and need help in align the output in a proper view format.
Tried code :
from AnilistPython import Anilist
import pandas as pd
import re
# db access online
anilist = Anilist()
# User input
ani_search = anilist.get_anime(input('Enter the Anime Name\t:\t'), manual_select=True)
df = ani_search
# for Genres split
cate = []
for gen in df['genres']:
cate.append(gen)
cate1 = (' , '.join(cate))
# for Checking Episode
if df['airing_status'] == 'RELEASING':
print('Ongoing')
x ='Ongoing'
y = df['next_airing_ep']
print(y['episode'])
y1 = y['episode']
elif df['airing_status'] == 'FINISHED':
print('Ended')
x = 'Ended'
y = df['airing_episodes']
print(y)
y1 = y
else:
print('None')
# print other details
print(f"\nTitle_Name\t:\t{df['name_english']}\nRomji_Title\t:\t{df['name_romaji']}\nPlot\t:\t{re.split('<br>', df['desc'])}\nAiring_Format\t:\t{df['airing_format']}\nStatus\t:\t{x}\nEpisodes_Count\t:\t{y1}\nGenres\t:\t{cate1}\nRating\t:\t{df['average_score']}/100\n")
The output it generated :
Enter the Anime Name : Bleach
1. BLEACH
2. BEACH
3. Akkanbee da
Please select the anime that you are searching for in number: 1
Title_Name : Bleach
Romji_Title : BLEACH
Plot : ["Ichigo Kurosaki is a rather normal high school student apart from the fact he has the ability to see ghosts. This ability never impacted his life in a major way until the day he encounters the Shinigami Kuchiki Rukia, who saves him and his family's lives from a Hollow, a corrupt spirit that devours human souls. \n", '', '\nWounded during the fight against the Hollow, Rukia chooses the only option available to defeat the monster and passes her Shinigami powers to Ichigo. Now forced to act as a substitute until Rukia recovers, Ichigo hunts down the Hollows that plague his town. \n\n\n']
Airing_Format : TV
Status : Ended
Episodes_Count : 366
Genres : Action , Adventure , Supernatural
Rating : 76/100
I am looking for the format to look like this:
Title_Name : Bleach
Romji_Title : BLEACH
Plot : Ichigo Kurosaki is a rather normal high school student apart from the fact he has the ability to see ghosts. This ability never impacted his life in a major way until the day he encounters the Shinigami Kuchiki Rukia, who saves him and his family's lives from a Hollow, a corrupt spirit that devours human souls. Wounded during the fight against the Hollow, Rukia chooses the only option available to defeat the monster and passes her Shinigami powers to Ichigo. Now forced to act as a substitute until Rukia recovers, Ichigo hunts down the Hollows that plague his town.
Airing_Format : TV
Status : Ended
Episodes_Count : 366
Genres : Action , Adventure , Supernatural
Rating : 76/100
Handling Expection
The full question is of Hackerrank. I have passed all the test cases but one test case is failing. I don't know why. Logic is correct. Please help me
test case in given question
*
4500(membership fee)
3(number of installments)
Welcome boy(name)
*
my code is
def Library(memberfee, installment, book):
if(installment > 3):
print("Maximum Permitted Number of Installments is 3")
else:
if(installment == 0):
print("Number of Installments cannot be Zero.")
else:
print("Amount per Installment is {}".format(memberfee/installment))
ListOfBooks = ["philosophers stone", "chamber of sec rets", "prisoner of azkaban", "goblet of fire", "order of phoenix", "half blood price", "deathly hallows 1", "deathly hallows2"]
book = book.lower()
if book in ListOfBooks:
print("It is available in this section")
else:
print("No such book exists in this section")
if __name__ == '__main__':
memberfee = int(input())
installment = int(input())
book = input()
try:
Library(memberfee,installment,book)
except ZeroDivisionError as e:
print(e)
except ValueError as e:
print(e)
except NameError as e:
print(e)
**i think the problem may be here
print("Amount per Installment is {}".format(memberfee/installment))
**
The question is.....
*Library
This exception handling scenario deals with the exceptional cases that arise in a typical library interface of a Town library.
O
1
About the Library Interface
This is a typical interface provided in the library, which takes 3 inputs from the library members, sequentially. They are:
memberfee - Membership fee for the library for the next financial year, which can be paid in installments.
installment - Number of installments chosen to pay the Membership fee. 3. book- Name of the book the member looks for in the 'Harry Potter' Section.
23
24
25
26
27
28
29
30
Note
All the above inputs except 'book' are Integers.
Write the function definition as follows, for the function 'Library', that takes all the above 3 inputs as its parameters:
The maximum permitted number of installments to pay the annual membership fee is '3'.
Raise ValueError exception if the input for the number of installments is greater than '3' and Print a Message. The message to the user must be, "Maximum Permitted Number of Installments is 3", The amount per installment is calculated by dividing the Membership fee by the number of installments. 2. Raise ZeroDivision Error exception if the input for the number of installments is equal to '0' and Print a Message. The message to the user must be, "Number of Installments cannot be Zero." else
1
#!/bin/python
10
11
def Library (
if(insta
prin
else:
if(
els
(memberfee
sec rets",
phoenix",
hallows 2"1
Print the amount per installment as "Amount per Installment is 3000.0".
12
13
The 'Harry Potter' book section contains the following books only:
14
ALL
1
15
philosophers stone
16
. chamber of secrets
17
• prisoner of azkaban
18
19
• goblet of fire
• order of phoenix
half blood prince
deathly hallows 1
21
23
20
24
• deathly hallows 2*
def Library(memberfee,installment,book):
# Write your code here
#print(memberfee)
#print(installment)
#print(book)
if installment > 3:
raise(ValueError("Maximum Permitted Number of Installments is 3"))
if installment == 0:
raise(ZeroDivisionError("Number of Installments cannot be Zero."))
else:
print ("Amount per Installment is ", memberfee / installment)
if book == 'philosophers stone' or book == 'Chamber of Secrets' or book == 'prisoner of azkaban' or book == 'Goblet of Fire' or book == 'order of phoenix' or book == 'Half Blood Prince' or book == 'Deathly Hallows 1' or book == 'deathly hallows 2':
print ("It is available in this section")
else:
raise(NameError("No such book exists in this section"))
Your ListofBooks contains some grammatical errors in the names of the books. Maybe that's what caused the error.
In my opinion, it is just a name not matching with listofbooks.
Please try to check all the book names from listofbooks.
I'm working on this side project game to grasp python better. I'm trying to have the user enter the amount of time the character has to spend busy, then not allow the user to do the same thing until they have completed the original time entered. I have tried a few methods with varying error results from my noob ways. (timestamps, converting input to int and time in different spots, timeDelta)
def Gold_mining():
while P.notMining:
print('Welcome to the Crystal mines kid.\nYou will be paid in gold for your labour,\nif lucky you may get some skill points or bonus finds...\nGoodluck in there.')
print('How long do you wish to enter for?')
time_mining = int(input("10 Gold Per hour. Max 8 hours --> "))
if time_mining > 0 and time_mining <= 8:
time_started = current_time
print(f'You will spend {time_mining} hours digging in the mines.')
P.gold += time_mining * 10
print(P.gold)
P.notMining = False
End_Time = (current_time + timedelta(hours = 2))
print(f'{End_Time} time you exit the mines...')
elif time_mining > 8:
print("You can't possibly mine for that long kid, go back and think about it.")
else:
print('Invalid')
After the set amount of time i would like for it to change the bool value back to false so that you can mine again.
"Crystal Mining" is mapped to a different key for testing so my output says "Inventory" but would say "Crystal Mining" when it works properly and currently looks like this:
*** Page One ***
Intro Page
02:15:05
1 Character Stats
2 Rename Character
3 Inventory
4 Change Element
5 Menu
6 Exit
Num: 3
Welcome to the Crystal mines kid.
You will be paid in gold for your labour,
if lucky you may get some skill points or bonus finds...
Goodluck in there.
How long do you wish to enter for?
10 Gold Per hour. Max 8 hours --> 1
You will spend 1 hours digging in the mines.
60
Traceback (most recent call last):
File "H:\Python ideas\input_as_always.py", line 176, in <module>
intro.pageInput()
File "H:\Python ideas\input_as_always.py", line 45, in pageInput
self.pageOptions[pInput]['entry']()
File "H:\Python ideas\input_as_always.py", line 134, in Gold_mining
End_Time = (current_time + timedelta(hours = 2))
TypeError: can only concatenate str (not "datetime.timedelta") to str
I have gotten a very strange data. I have dictionary with keys and values where I want to use this dictionary to search if these keywords are ONLY starting and/or end of the text not middle of the sentence. I tried to create simple data frame below to show the problem case and python codes that I have tried so far. How do I get it go search for only starting or ending of the sentence? This one searches whole text sub-strings.
Code:
d = {'apple corp':'Company','app':'Application'} #dictionary
l1 = [1, 2, 3,4]
l2 = [
"The word Apple is commonly confused with Apple Corp which is a business",
"Apple Corp is a business they make computers",
"Apple Corp also writes App",
"The Apple Corp also writes App"
]
df = pd.DataFrame({'id':l1,'text':l2})
df['text'] = df['text'].str.lower()
df
Original Dataframe:
id text
1 The word Apple is commonly confused with Apple Corp which is a business
2 Apple Corp is a business they make computers
3 Apple Corp also writes App
4 The Apple Corp also writes App
Code Tried out:
def matcher(k):
x = (i for i in d if i in k)
# i.startswith(k) getting error
return ';'.join(map(d.get, x))
df['text_value'] = df['text'].map(matcher)
df
Error:
TypeError: 'in <string>' requires string as left operand, not bool
when I use this x = (i for i in d if i.startswith(k) in k)
Empty values if i tried this x = (i for i in d if i.startswith(k) == True in k)
TypeError: sequence item 0: expected str instance, NoneType found
when i use this x = (i.startswith(k) for i in d if i in k)
Results from Code above ... Create new field 'text_value':
id text text_value
1 The word Apple is commonly confused with Apple Corp which is a business Company;Application
2 Apple Corp is a business they make computers Company;Application
3 Apple Corp also writes App Company;Application
4 The Apple Corp also writes App Company;Application
Trying to get an FINAL output like this:
id text text_value
1 The word Apple is commonly confused with Apple Corp which is a business NaN
2 Apple Corp is a business they make computers Company
3 Apple Corp also writes App Company;Application
4 The Apple Corp also writes App Application
You need a matcher function which can accept flag and then call that twice to get the results for startswith and endswith.
def matcher(s, flag="start"):
if flag=="start":
for i in d:
if s.startswith(i):
return d[i]
else:
for i in d:
if s.endswith(i):
return d[i]
return None
df['st'] = df['text'].apply(matcher)
df['ed'] = df['text'].apply(matcher, flag="end")
df['text_value'] = df[['st', 'ed']].apply(lambda x: ';'.join(x.dropna()),1)
df = df[['id','text', 'text_value']]
The text_value column looks like:
0
1 Company
2 Company;Application
3 Application
Name: text_value, dtype: object
joined = "|".join(d.keys())
pat = '(?i)^(?:the\\s*)?(' + joined + ')\\b.*?|.*\\b(' + joined + ')$'+'|.*'
get = lambda x: d.get(x.group(1),"") + (';' +d.get(x.group(2),"") if x.group(2) else '')
df.text.str.replace(pat,get)
0
1 Company
2 Company;Application
3 Company;Application
Name: text, dtype: object
I want to color the interior of a cell according to it's content, however when I'm accessing its value I'm always getting '1.0', the value is calculated.
Colorisation code :
def _colorizeTop10RejetsSheet(self):
"""Colore les position de la page "Top 10 Rejets" """
start_position = (5, 12)
last_line = 47
for x in range(start_position[0], last_line+1):
current_cell = self.workbook.Sheets("Top 10 Rejets").Cells(x, start_position[1])
current_cell.Interior.Color = self._computePositionColor(current_cell.Value)
def _computePositionColor(self, position):
"""Colore les position de 1 a 5 en rouge de et 6 a 10 en orange"""
if position < 6:
return self.RED
elif position < 11:
return self.ORANGE
else:
return self.WHITE
Excel cell code :
=SI(ESTNA(RECHERCHEV(CONCATENER(TEXTE($F23;0);TEXTE($G23;"00");$H23;$I23);Données!$J:$P;7;FAUX));MAX(Données!$P:$P);RECHERCHEV(CONCATENER(TEXTE($F23;0);TEXTE($G23;"00");$H23;$I23);Données!$J:$P;7;FAUX))
How could I get the calculated value?
I'm using python 2.7 and I'm communicating with Excel through win32com
Thanks
Adding this to the beginning of the _colorizeTop10Rejets method did the trick
self.xl.Calculate()
self.xl is the object returned by win32.Dispatch('Excel.Application')