Similar to the question that was asked previously amazon lex transcript wide open slot. This dealing with the solution to that question and the comment under the solution. The intent I want to create has only one slot and I just want to respond appropriately after receiving the reply from the user. in this case, if I don't want to use elicitSlot. What should I do?
I have got the first part down:
slots = intent_request['currentIntent']['slots']
slots['anyString'] = intent_request['inputTranscript']
I tried to use elixit_slot. But it will only prompt me for the same transcript again, which is not what I want.
I want to output a reply back to lex after the user entered a set of strings after the first time the inputTranscript is filled up.
I have a workaround for this problem:
if ChooseCase == 'Case 1' and CaseOneChosen == 'Yes' and not CaseOneQuestion:
# Assign current inputTranscript to slots
intent_request['currentIntent']['slots']['CaseOneQuestion'] = intent_request['inputTranscript']
CaseOneQuestion = intent_request['inputTranscript']
# After the user enter the text, check if the current slot value is the same as the previous slot,
# if it is, then this is the first time the user elicit this slot
# Else, assign the current inputTranscript to the current slot (done above)
if intent_request['currentIntent']['slots']['CaseOneQuestion'] == intent_request['currentIntent']['slots']['CaseOneChosen']:
return elicit_slot(
output_session_attributes,
intent_request['currentIntent']['name'],
intent_request['currentIntent']['slots'],
'CaseOneQuestion',
{'contentType': 'PlainText', 'content': 'Answer for case 1'},
None
)
Explain: The previous slot is "CaseOneChosen", the value is this slot is "Yes", then line 3 will assign that value to the slot "CaseOneQuestion". Now they both have the same value, so the if statement is TRUE. The bot will let the user elicit slot "CaseOneQuestion" (I assume the user entered "cat"). After this, line 3 will assign "cat" to "CaseOneQuestion" slot. So now the if statement is FALSE. the bot will not re-prompt to ask the user elicit "CaseOneQuestion" slot anymore. And the slot value is now "cat".
Hope this helps.
Related
I am using user_data[TO] = update.message.text to update variables as a user types information into the bot.
When it comes time to output those variables, it is only showing sequential numbers starting at 0.
FROM, TO, SEND, REVIEW = range(4)
user_data = {}
I am using this to store the variables.
and this to output them.
def review_handler(update: Update, context):
update.message.reply_text('*Review your Details* \n\n'
f'Data1 \+{FROM} \n'
f'Data2 \+{TO} \n\n'
'Press /yes to proceed and /cancel to exit', parse_mode='MarkdownV2')
I am getting output like Data1 +0 Data2 +1
and so on. Not sure what the issue is.
I figured it out. I needed to use {user_data[TO]} instead of just [TO]
I'm working on trying to add an array formula so that the status column updates as information is entered.
This is an order tracking sheet. When an order is entered get set to Available, Once a driver is assigned change the status to Dispatched. As times are entered in the In and Out columns to change from Picked to Delivered. Finally once its checked off as Billed to mark the status as Complete.
So far I've only gotten to
=ARRAYFORMULA(IF(LEN(D3:D&H3:H&I3:I&M3:M&N3:N)=0, "Available", IF(D3:D<>"", "Dispatched")))
I haven't been able to figure out past that.
https://docs.google.com/spreadsheets/d/13tOkLwtPYyWm9rUfkCidqwUygzFfVfmIzV-EwL5i7hM/edit?usp=sharing
I wasn't sure which column was used to mark billed, so I just guess column P, just change it to the proper column. Enter this formula in row 1 of your status column.
={"Status";
ARRAYFORMULA(
IFERROR(
IFS(
P2:P<>"", "Complete",
J2:J<>"", "Delivered",
I2:I<>"", "Picked",
E2:E<>"", "Dispatched"
),
"")
)}
Explanation
Starting inside and working out:
The ranges J2:J, etc, tell the array formula which column to evaluate as it goes row by row, starting at row 2, and ending at the bottom of the sheet.
The IFS formula checks condition:result pairs, If the first condition is met, it displays that value, but if not, it goes to the next one, and so on until it finds a condition that is true or reaches the end.
The IFS formula will return an error if it doesn't find any of the conditions to be true, so the IFERROR says to leave the cell blank if there are no true conditions.
The ARRAYFORMULA evaluates each row for the defined range.
The { xxxx ; xxxxxxx}stacks the items after the semicolon below the item before the semicolon. In this usage, the label "Status" is in row 1, and the results are in the rows below it. This ensures that no one accidentally sorts or erases the formula.
I am trying to distinguish flight numbers.
Example:
flightno = "FR556"
split_data = flightno.upcase.match(/([A-Za-z]+)(\d+)/)
first = split_data[1] # FR
second = split_data[1] # 556
I then go on to query the database to find an airline based on the FR in this example and apply some logic with the result which is Ryanair.
My problem is when the flight number might be:
flightno = "U21920"
split_data = flightno.upcase.match(/([A-Za-z]+)(\d+)/)
first = split_data[1] # U
second = split_data[1] # 21920
i basically want first to be U2 not just U. This is used to search the database of airlines by their IATA code in this case is U2
****EDIT**
In the interest of clarity i made some mistakes in terminology when asking my question. Due to the complexities of booking reference numbers, the input is taken from whatever the passenger provides. For an easyJet flight for example, the passenger may input EZY1920 or U21920 only the airline provides either so the passenger is ignorant really.
"EZY" = ICAO
"U2" = IATA
I take the input from the user and try to separate the ICAO or IATA from the flight number "1920" but there is no way of determining that without searching the database or separating the input which i feel is cumbersome from a user experience point of view.
Using a regex to separate characters from numbers works until the user inputs an IATA as part of their flight number (the passenger won't know the difference) and as you can see in the example above this confuses the regex.**
The trouble is i cant think of any other pattern with flight numbers. They always have at least two characters made up of just letters or a mixture of a letter and a number and can be 3 characters in length. The numbers part can be as short as 1 but can also be as long as 4 - always numbers.
****edit**
As has been mentioned in the comments, there is no fixed size however one thing that is always true (at least so far) is the first character will always be a letter regardless if it is ICAO or IATA.
After considering every bodies input so far i'm wondering if searching the database and returning airlines with an IATA or ICAO that matches the first two letters provided by the user (U2), (FR), (EZ) might be one way to go, however this is subject to obvious problems should an ICAO or IATA be released that matches another airline, for example "EZY" & "EZT". This is not future proof and i'm looking for better ruby or regex solutions.**
Appreciate your input.
EDIT
I have answered my own question below. While other answers provide a solution for handling some conditions they would fall down if the flight number began with a number so i worked out a crass but to date stable way to analyse the string for digits and then work out if it is an ICAO or IATA from that.
A solution I think of is that you match your given flight number against a complete list of ICAO/IATA codes: https://raw.githubusercontent.com/datasets/airport-codes/master/data/airport-codes.csv
Spending some time with google might give you a more appropriate list.
Then use the first three characters (if that is the maximum) of your flight number to find a match within the icao codes. If you find one, you will know where to seperate your string.
Here a minimal ugly example that should set you on a track. Feel free to update!
ICAOCODES = %w(FR DEU U21) # grab your data here
def retrieve_flight_information(flightnumber)
ICAOCODES.each do |icao|
co = flightnumber.match(icao).to_s
if co.length > 0
# airline
puts co
# flight number
puts flightnumber.gsub(co,'')
end
end
end
retrieve_flight_information("FR556")
#=> FR
#=> 556
retrieve_flight_information("U21214123")
#=> U21
#=> 214123
The biggest flaw lies in using .gsub() as it might mess up your flightnumber in case it looks like this: "FR21413FR2"
However you will find plenty of solutions to this problem on so.
As mentioned in the comments, a list of icao codes is not what you are looking for. But what is relevant here, is that you somehow need a list of strings that you can securely compare against.
I have a fairly crass solution that seems to be working in all scenarios i can throw at it to date. I wanted to make this available to anybody else that might find it useful?
The general rule of thumb for flight codes/numbers seems to be:
IATA: two characters made up of any combination letters and digits
ICAO: three characters made up of letters only (to date)
With that in mind we should be able to work out if we need to search the database by IATA or ICAO depending on the condition of the first three characters.
First we take the flight number and convert to uppercase
string = "U21920".upcase
Next we analyse the first three characters to check for any numbers.
first_three = string[0,3] # => U21
Is there a digit in first_three?
if first_three =~ /\d/ # => true
iata = first_three[0,2] # => If true lets get rid of the last character
# Now we go to the database searching IATA (U2)
search = Airline.where('iata LIKE ?', "#{iata}%") # => Starts with search, just in case
Otherwise if there isnt a digit found in the string
else
icao = string.match(/([A-Za-z]+)(\d+)/)
search = Airline.where('icao LIKE ?', "#{icao[1]}%")
This seems to work for the random flight numbers ive tested it with today from a few of the major airport live departure/arrival boards. Its an interesting problem because some airlines issue tickets with either an ICAO or IATA code as part of the flight number which means passengers won't know any different, not to mention, some airports provide flight information in their own format so assumign there isnt a change to the ICAO and IATA build then the above should work.
Here is an example script you can run
test.rb
puts "What is your flight number?"
string = gets.upcase
first_three = string[0,3]
puts "Taking first three from #{string} is #{first_three}"
if first_three =~ /\d/ # Calling String's =~ method.
puts "The String #{first_three} DOES have a number in it."
iata = first_three[0,2]
search = Airline.where('iata LIKE ?', "#{iata}%")
puts "Searching Airlines starting with IATA #{iata} = #{search.count}"
puts "Found #{search.first.name} from IATA #{iata}"
else
puts "The String #{first_three} does not have a number in it."
icao = string.match(/([A-Za-z]+)(\d+)/)
search = Airline.where('icao LIKE ?', "#{icao[1]}%")
puts "Searching Airlines starting with ICAO #{icao[1]} = #{search.count}"
puts "Found #{search.first.name} from IATA #{icao[1]}"
end
Airline
Airline(id: integer, name: string, iata: string, icao: string, created_at: datetime, updated_at: datetime )
stick this in your lib folder and run
rails runner lib/test.rb
Obviously you can remove all of the puts statements to get straight to the result. I'm using rails runner to include access to my Airline model when running the script.
it run corectly but it should have around 500 matches but it only has around 50 and I dont know why!
This is a probelm for my comsci class that I am having isues with
we had to make a function that checks a list for duplication I got that part but then we had to apply it to the birthday paradox( more info here http://en.wikipedia.org/wiki/Birthday_problem) thats where I am runing into problem because my teacher said that the total number of times should be around 500 or 50% but for me its only going around 50-70 times or 5%
duplicateNumber=0
import random
def has_duplicates(listToCheck):
for i in listToCheck:
x=listToCheck.index(i)
del listToCheck[x]
if i in listToCheck:
return True
else:
return False
listA=[1,2,3,4]
listB=[1,2,3,1]
#print has_duplicates(listA)
#print has_duplicates(listB)
for i in range(0,1000):
birthdayList=[]
for i in range(0,23):
birthday=random.randint(1,365)
birthdayList.append(birthday)
x= has_duplicates(birthdayList)
if x==True:
duplicateNumber+=1
else:
pass
print "after 1000 simulations with 23 students there were", duplicateNumber,"simulations with atleast one match. The approximate probibilatiy is", round(((duplicateNumber/1000)*100),3),"%"
This code gave me a result in line with what you were expecting:
import random
duplicateNumber=0
def has_duplicates(listToCheck):
number_set = set(listToCheck)
if len(number_set) is not len(listToCheck):
return True
else:
return False
for i in range(0,1000):
birthdayList=[]
for j in range(0,23):
birthday=random.randint(1,365)
birthdayList.append(birthday)
x = has_duplicates(birthdayList)
if x==True:
duplicateNumber+=1
print "after 1000 simulations with 23 students there were", duplicateNumber,"simulations with atleast one match. The approximate probibilatiy is", round(((duplicateNumber/1000.0)*100),3),"%"
The first change I made was tidying up the indices you were using in those nested for loops. You'll see I changed the second one to j, as they were previously bot i.
The big one, though, was to the has_duplicates function. The basic principle here is that creating a set out of the incoming list gets the unique values in the list. By comparing the number of items in the number_set to the number in listToCheck we can judge whether there are any duplicates or not.
Here is what you are looking for. As this is not standard practice (to just throw code at a new user), I apologize if this offends any other users. However, I believe showing the OP a correct way to write a program should be could all do us a favor if said user keeps the lack of documentation further on in his career.
Thus, please take a careful look at the code, and fill in the blanks. Look up the python doumentation (as dry as it is), and try to understand the things that you don't get right away. Even if you understand something just by the name, it would still be wise to see what is actually happening when some built-in method is being used.
Last, but not least, take a look at this code, and take a look at your code. Note the differences, and keep trying to write your code from scratch (without looking at mine), and if it messes up, see where you went wrong, and start over. This sort of practice is key if you wish to succeed later on in programming!
def same_birthdays():
import random
'''
This is a program that does ________. It is really important
that we tell readers of this code what it does, so that the
reader doesn't have to piece all of the puzzles together,
while the key is right there, in the mind of the programmer.
'''
count = 0
#Count is going to store the number of times that we have the same birthdays
timesToRun = 1000 #timesToRun should probably be in a parameter
#timesToRun is clearly defined in its name as well. Further elaboration
#on its purpose is not necessary.
for i in range(0,timesToRun):
birthdayList = []
for j in range(0,23):
random_birthday = random.randint(1,365)
birthdayList.append(random_birthday)
birthdayList = sorted(birthdayList) #sorting for easier matching
#If we really want to, we could provide a check in the above nester
#for loop to check right away if there is a duplicate.
#But again, we are here
for j in range(0, len(birthdayList)-1):
if (birthdayList[j] == birthdayList[j+1]):
count+=1
break #leaving this nested for-loop
return count
If you wish to find the percent, then get rid of the above return statement and add:
return (count/timesToRun)
Here's a solution that doesn't use set(). It also takes a different approach with the array so that each index represents a day of the year. I also removed the hasDuplicate() function.
import random
sim_total=0
birthdayList=[]
#initialize an array of 0's representing each calendar day
for i in range(365):
birthdayList.append(0)
for i in range(0,1000):
first_dup=True
for n in range(365):
birthdayList[n]=0
for b in range(0, 23):
r = random.randint(0,364)
birthdayList[r]+=1
if (birthdayList[r] > 1) and (first_dup==True):
sim_total+=1
first_dup=False
avg = float(sim_total) / 1000 * 100
print "after 1000 simulations with 23 students there were", sim_total,"simulations with atleast one duplicate. The approximate problibility is", round(avg,3),"%"
I need help with the following H.W. problem. I have done everything except the instructions I numbered. Please help!
A furniture manufacturer makes two types of furniture—chairs and sofas.
The cost per chair is $350, the cost per sofa is $925, and the sales tax rate is 5%.
Write a Visual Basic program to create an invoice form for an order.
After the data on the left side of the form are entered, the user can display an invoice in a list box by pressing the Process Order button.
The user can click on the Clear Order Form button to clear all text boxes and the list box, and can click on the Quit button to exit the program.
The invoice number consists of the capitalized first two letters of the customer’s last name, followed by the last four digits of the zip code.
The customer name is input with the last name first, followed by a comma, a space, and the first name. However, the name is displayed in the invoice in the proper order.
The generation of the invoice number and the reordering of the first and last names should be carried out by Function procedures.
Seeing as this is homework and you haven't provided any code to show what effort you have made on your own, I'm not going to provide any specific answers, but hopefully I will try to point you in the right direction.
Your first 2 numbered items look to be variations on the same theme... string manipulation. Assuming you have the customer's address information from the order form, you just need to write 2 separate function to take the parts of the name and address, take the data you need and return the value (which covers your 3rd item).
To get parts of the name and address to generate the invoice number, you need to think about using the Left() and Right() functions.
Something like:
Dim first as String, last as String, word as String
word = "Foo"
first = Left(word, 1)
last = Right(word, 1)
Debug.Print(first) 'prints "F"
Debug.Print(last) 'prints "o"
Once you get the parts you need, then you just need to worry about joining the parts together in the order you want. The concatenation operator for strings is &. So using the above example, it would go something like:
Dim concat as String
concat = first & last
Debug.Print(concat) 'prints "Fo"
Your final item, using a Function procedure to generate the desired values, is very easily google-able (is that even a word). The syntax is very simple, so here's a quick example of a common function that is not built into VB6:
Private Function IsOdd(value as Integer) As Boolean
If (value Mod 2) = 0 Then 'determines of value is an odd or even by checking
' if the value divided by 2 has a remainder or not
' (aka Mod operator)
IsOdd = False ' if remainder is 0, set IsOdd to False
Else
IsOdd = True ' otherwise set IsOdd to True
End If
End Function
Hopefully this gets you going in the right direction.