Hello,

Is there a way in Architect to COUNT in reverse? For example, if I have an array of {Apple, Pear, Banana, Banana} and I want to use the COUNT function but start from the right, excluding anything Banana, how would I go about that? I considered using the REVERSE function but do not want to reverse the actual words, just the order I'm counting them in.

Regards,
Mo
#ArchitectureandDesign
#Routing(ACD/IVR)

------------------------------
Mo Ford
Upgrade
------------------------------

Use a loop to go through each item in your collection.  In this example Task.fruit is the string collection.  Name the index variable anything you like.  The maximum loop count tells the loop when to stop, in your case you want the number of items in your list, so use the expression Count(Task.fruit) which will be 4.  Normally people want to go through the list from the start, so inside the loop you would use Task.fruit[Task.index].  You know collection indices start at 0 right?  So
0=Apple
1=Pear
2=Banana
3=Banana
In your case you want to go through it backwards, so instead of Task.fruit[Task.index] you want to look at Task.fruit[Count(Task.fruit)-1-Task.index]. 
When Task.index=0, you will look at the item with index 4-1-0 = 3 and get Banana
When Task.index=1, you will look at the item with index 4-1-1 = 2 and get Banana
When Task.index=2, you will look at the item with index 4-1-2 = 1 and get Pear
When Task.index=3, you will look at the item with index 4-1-3 = 0 and get Apple

FYI there's a much simpler way of counting how many items are not Banana in your collection.  You could just evaluate this expression, no looping necessary.
Count(Task.fruit)-Count(Find(Task.fruit, "Banana"))
Find returns a collection of indices for which the item exists in the given collection (in your case you would get a collection with the values 2 & 3).  The count of that is 2, and subtracting it from the total count for the collection gets 2.

------------------------------
Melissa Bailey
Genesys - Employees
------------------------------

Original Message:
Sent: 01-25-2022 06:00
From: Mo Ford
Subject: Count in Reverse Order

Hello,

Is there a way in Architect to COUNT in reverse? For example, if I have an array of {Apple, Pear, Banana, Banana} and I want to use the COUNT function but start from the right, excluding anything Banana, how would I go about that? I considered using the REVERSE function but do not want to reverse the actual words, just the order I'm counting them in.

Regards,
Mo
#ArchitectureandDesign
#Routing(ACD/IVR)

------------------------------
Mo Ford
Upgrade
------------------------------

Thank you, Melissa. The second suggestion is much cleaner and worked for finding the count.

If I wanted to instead find the location of the items in the string below that are not EXPIRED, CLOSED or REJECTED with the intent to save the location to be used later in a GetAt function, how would I go about doing that?

EXPIRED, ISSUED, CLOSED, OPENED, REJECTED

------------------------------
Mo Ford
Upgrade
------------------------------

Original Message:
Sent: 01-25-2022 09:46
From: Melissa Bailey
Subject: Count in Reverse Order

Use a loop to go through each item in your collection.  In this example Task.fruit is the string collection.  Name the index variable anything you like.  The maximum loop count tells the loop when to stop, in your case you want the number of items in your list, so use the expression Count(Task.fruit) which will be 4.  Normally people want to go through the list from the start, so inside the loop you would use Task.fruit[Task.index].  You know collection indices start at 0 right?  So
0=Apple
1=Pear
2=Banana
3=Banana
In your case you want to go through it backwards, so instead of Task.fruit[Task.index] you want to look at Task.fruit[Count(Task.fruit)-1-Task.index].
When Task.index=0, you will look at the item with index 4-1-0 = 3 and get Banana
When Task.index=1, you will look at the item with index 4-1-1 = 2 and get Banana
When Task.index=2, you will look at the item with index 4-1-2 = 1 and get Pear
When Task.index=3, you will look at the item with index 4-1-3 = 0 and get Apple

FYI there's a much simpler way of counting how many items are not Banana in your collection.  You could just evaluate this expression, no looping necessary.
Count(Task.fruit)-Count(Find(Task.fruit, "Banana"))
Find returns a collection of indices for which the item exists in the given collection (in your case you would get a collection with the values 2 & 3).  The count of that is 2, and subtracting it from the total count for the collection gets 2.

------------------------------
Melissa Bailey
Genesys - Employees

Original Message:
Sent: 01-25-2022 06:00
From: Mo Ford
Subject: Count in Reverse Order

Hello,

Is there a way in Architect to COUNT in reverse? For example, if I have an array of {Apple, Pear, Banana, Banana} and I want to use the COUNT function but start from the right, excluding anything Banana, how would I go about that? I considered using the REVERSE function but do not want to reverse the actual words, just the order I'm counting them in.

Regards,
Mo
#ArchitectureandDesign
#Routing(ACD/IVR)

------------------------------
Mo Ford
Upgrade
------------------------------

First create a string collection Task.statuses from your string using the Split function.  Create an integer collection Task.GoodStatusIndices to hold the indices (locations) of the items that don't match.  Create another string collection Task.BadStatuses with EXPIRED, CLOSED, & REJECTED as the items.  Then use a loop to go through each item in Task.statuses and determine if it is not in Task.BadStatuses.  If so, add the index to Task.GoodStatusIndices.

The full expression in the decision action is 
FindFirst(Task.BadStatuses, Trim(Task.statuses[Task.index])) == -1
Trim is getting rid of any leading or trailing whitespace.  FindFirst returns -1 if the item is not found anywhere in the collection.
The full expression for the update data action inside the loop is
AddItem(Task.GoodStatusIndices, Task.index)




------------------------------
Melissa Bailey
Genesys - Employees
------------------------------

Original Message:
Sent: 01-27-2022 10:36
From: Mo Ford
Subject: Count in Reverse Order

Thank you, Melissa. The second suggestion is much cleaner and worked for finding the count.

If I wanted to instead find the location of the items in the string below that are not EXPIRED, CLOSED or REJECTED with the intent to save the location to be used later in a GetAt function, how would I go about doing that?

EXPIRED, ISSUED, CLOSED, OPENED, REJECTED

------------------------------
Mo Ford
Upgrade

Original Message:
Sent: 01-25-2022 09:46
From: Melissa Bailey
Subject: Count in Reverse Order

Use a loop to go through each item in your collection.  In this example Task.fruit is the string collection.  Name the index variable anything you like.  The maximum loop count tells the loop when to stop, in your case you want the number of items in your list, so use the expression Count(Task.fruit) which will be 4.  Normally people want to go through the list from the start, so inside the loop you would use Task.fruit[Task.index].  You know collection indices start at 0 right?  So
0=Apple
1=Pear
2=Banana
3=Banana
In your case you want to go through it backwards, so instead of Task.fruit[Task.index] you want to look at Task.fruit[Count(Task.fruit)-1-Task.index].
When Task.index=0, you will look at the item with index 4-1-0 = 3 and get Banana
When Task.index=1, you will look at the item with index 4-1-1 = 2 and get Banana
When Task.index=2, you will look at the item with index 4-1-2 = 1 and get Pear
When Task.index=3, you will look at the item with index 4-1-3 = 0 and get Apple

FYI there's a much simpler way of counting how many items are not Banana in your collection.  You could just evaluate this expression, no looping necessary.
Count(Task.fruit)-Count(Find(Task.fruit, "Banana"))
Find returns a collection of indices for which the item exists in the given collection (in your case you would get a collection with the values 2 & 3).  The count of that is 2, and subtracting it from the total count for the collection gets 2.

------------------------------
Melissa Bailey
Genesys - Employees

Original Message:
Sent: 01-25-2022 06:00
From: Mo Ford
Subject: Count in Reverse Order

Hello,

Is there a way in Architect to COUNT in reverse? For example, if I have an array of {Apple, Pear, Banana, Banana} and I want to use the COUNT function but start from the right, excluding anything Banana, how would I go about that? I considered using the REVERSE function but do not want to reverse the actual words, just the order I'm counting them in.

Regards,
Mo
#ArchitectureandDesign
#Routing(ACD/IVR)

------------------------------
Mo Ford
Upgrade
------------------------------

Thanks, Melissa. That suggestion worked to gather the locations of the good statuses. If I wanted to reference the highest number in the good status indices, how would I go about doing that in a GetAt() function?

------------------------------
Mo Ford
Upgrade
------------------------------

Original Message:
Sent: 01-31-2022 13:10
From: Melissa Bailey
Subject: Count in Reverse Order

First create a string collection Task.statuses from your string using the Split function.  Create an integer collection Task.GoodStatusIndices to hold the indices (locations) of the items that don't match.  Create another string collection Task.BadStatuses with EXPIRED, CLOSED, & REJECTED as the items.  Then use a loop to go through each item in Task.statuses and determine if it is not in Task.BadStatuses.  If so, add the index to Task.GoodStatusIndices.

The full expression in the decision action is
FindFirst(Task.BadStatuses, Trim(Task.statuses[Task.index])) == -1
Trim is getting rid of any leading or trailing whitespace.  FindFirst returns -1 if the item is not found anywhere in the collection.
The full expression for the update data action inside the loop is
AddItem(Task.GoodStatusIndices, Task.index)




------------------------------
Melissa Bailey
Genesys - Employees

Original Message:
Sent: 01-27-2022 10:36
From: Mo Ford
Subject: Count in Reverse Order

Thank you, Melissa. The second suggestion is much cleaner and worked for finding the count.

If I wanted to instead find the location of the items in the string below that are not EXPIRED, CLOSED or REJECTED with the intent to save the location to be used later in a GetAt function, how would I go about doing that?

EXPIRED, ISSUED, CLOSED, OPENED, REJECTED

------------------------------
Mo Ford
Upgrade

Original Message:
Sent: 01-25-2022 09:46
From: Melissa Bailey
Subject: Count in Reverse Order

Use a loop to go through each item in your collection.  In this example Task.fruit is the string collection.  Name the index variable anything you like.  The maximum loop count tells the loop when to stop, in your case you want the number of items in your list, so use the expression Count(Task.fruit) which will be 4.  Normally people want to go through the list from the start, so inside the loop you would use Task.fruit[Task.index].  You know collection indices start at 0 right?  So
0=Apple
1=Pear
2=Banana
3=Banana
In your case you want to go through it backwards, so instead of Task.fruit[Task.index] you want to look at Task.fruit[Count(Task.fruit)-1-Task.index].
When Task.index=0, you will look at the item with index 4-1-0 = 3 and get Banana
When Task.index=1, you will look at the item with index 4-1-1 = 2 and get Banana
When Task.index=2, you will look at the item with index 4-1-2 = 1 and get Pear
When Task.index=3, you will look at the item with index 4-1-3 = 0 and get Apple

FYI there's a much simpler way of counting how many items are not Banana in your collection.  You could just evaluate this expression, no looping necessary.
Count(Task.fruit)-Count(Find(Task.fruit, "Banana"))
Find returns a collection of indices for which the item exists in the given collection (in your case you would get a collection with the values 2 & 3).  The count of that is 2, and subtracting it from the total count for the collection gets 2.

------------------------------
Melissa Bailey
Genesys - Employees

Original Message:
Sent: 01-25-2022 06:00
From: Mo Ford
Subject: Count in Reverse Order

Hello,

Is there a way in Architect to COUNT in reverse? For example, if I have an array of {Apple, Pear, Banana, Banana} and I want to use the COUNT function but start from the right, excluding anything Banana, how would I go about that? I considered using the REVERSE function but do not want to reverse the actual words, just the order I'm counting them in.

Regards,
Mo
#ArchitectureandDesign
#Routing(ACD/IVR)

------------------------------
Mo Ford
Upgrade
------------------------------

You mean get the last item?  

GetAt(Task.GoodStatusIndices, Count(Task.GoodStatusIndices)-1)​

------------------------------
Melissa Bailey
Genesys - Employees
------------------------------

Original Message:
Sent: 02-15-2022 04:54
From: Mo Ford
Subject: Count in Reverse Order

Thanks, Melissa. That suggestion worked to gather the locations of the good statuses. If I wanted to reference the highest number in the good status indices, how would I go about doing that in a GetAt() function?

------------------------------
Mo Ford
Upgrade

Original Message:
Sent: 01-31-2022 13:10
From: Melissa Bailey
Subject: Count in Reverse Order

First create a string collection Task.statuses from your string using the Split function.  Create an integer collection Task.GoodStatusIndices to hold the indices (locations) of the items that don't match.  Create another string collection Task.BadStatuses with EXPIRED, CLOSED, & REJECTED as the items.  Then use a loop to go through each item in Task.statuses and determine if it is not in Task.BadStatuses.  If so, add the index to Task.GoodStatusIndices.

The full expression in the decision action is
FindFirst(Task.BadStatuses, Trim(Task.statuses[Task.index])) == -1
Trim is getting rid of any leading or trailing whitespace.  FindFirst returns -1 if the item is not found anywhere in the collection.
The full expression for the update data action inside the loop is
AddItem(Task.GoodStatusIndices, Task.index)




------------------------------
Melissa Bailey
Genesys - Employees

Original Message:
Sent: 01-27-2022 10:36
From: Mo Ford
Subject: Count in Reverse Order

Thank you, Melissa. The second suggestion is much cleaner and worked for finding the count.

If I wanted to instead find the location of the items in the string below that are not EXPIRED, CLOSED or REJECTED with the intent to save the location to be used later in a GetAt function, how would I go about doing that?

EXPIRED, ISSUED, CLOSED, OPENED, REJECTED

------------------------------
Mo Ford
Upgrade

Original Message:
Sent: 01-25-2022 09:46
From: Melissa Bailey
Subject: Count in Reverse Order

Use a loop to go through each item in your collection.  In this example Task.fruit is the string collection.  Name the index variable anything you like.  The maximum loop count tells the loop when to stop, in your case you want the number of items in your list, so use the expression Count(Task.fruit) which will be 4.  Normally people want to go through the list from the start, so inside the loop you would use Task.fruit[Task.index].  You know collection indices start at 0 right?  So
0=Apple
1=Pear
2=Banana
3=Banana
In your case you want to go through it backwards, so instead of Task.fruit[Task.index] you want to look at Task.fruit[Count(Task.fruit)-1-Task.index].
When Task.index=0, you will look at the item with index 4-1-0 = 3 and get Banana
When Task.index=1, you will look at the item with index 4-1-1 = 2 and get Banana
When Task.index=2, you will look at the item with index 4-1-2 = 1 and get Pear
When Task.index=3, you will look at the item with index 4-1-3 = 0 and get Apple

FYI there's a much simpler way of counting how many items are not Banana in your collection.  You could just evaluate this expression, no looping necessary.
Count(Task.fruit)-Count(Find(Task.fruit, "Banana"))
Find returns a collection of indices for which the item exists in the given collection (in your case you would get a collection with the values 2 & 3).  The count of that is 2, and subtracting it from the total count for the collection gets 2.

------------------------------
Melissa Bailey
Genesys - Employees

Original Message:
Sent: 01-25-2022 06:00
From: Mo Ford
Subject: Count in Reverse Order

Hello,

Is there a way in Architect to COUNT in reverse? For example, if I have an array of {Apple, Pear, Banana, Banana} and I want to use the COUNT function but start from the right, excluding anything Banana, how would I go about that? I considered using the REVERSE function but do not want to reverse the actual words, just the order I'm counting them in.

Regards,
Mo
#ArchitectureandDesign
#Routing(ACD/IVR)

------------------------------
Mo Ford
Upgrade
------------------------------

Thanks for all of your help @Melissa Bailey! Last question (for now), if the collection is a collection of dates, is there a way to identify the most recent date? I assume some form of the date that's closest to today.​

------------------------------
Mo Ford
Upgrade
------------------------------

Original Message:
Sent: 02-17-2022 10:42
From: Melissa Bailey
Subject: Count in Reverse Order

You mean get the last item?

GetAt(Task.GoodStatusIndices, Count(Task.GoodStatusIndices)-1)​

------------------------------
Melissa Bailey
Genesys - Employees

Original Message:
Sent: 02-15-2022 04:54
From: Mo Ford
Subject: Count in Reverse Order

Thanks, Melissa. That suggestion worked to gather the locations of the good statuses. If I wanted to reference the highest number in the good status indices, how would I go about doing that in a GetAt() function?

------------------------------
Mo Ford
Upgrade

Original Message:
Sent: 01-31-2022 13:10
From: Melissa Bailey
Subject: Count in Reverse Order

First create a string collection Task.statuses from your string using the Split function.  Create an integer collection Task.GoodStatusIndices to hold the indices (locations) of the items that don't match.  Create another string collection Task.BadStatuses with EXPIRED, CLOSED, & REJECTED as the items.  Then use a loop to go through each item in Task.statuses and determine if it is not in Task.BadStatuses.  If so, add the index to Task.GoodStatusIndices.

The full expression in the decision action is
FindFirst(Task.BadStatuses, Trim(Task.statuses[Task.index])) == -1
Trim is getting rid of any leading or trailing whitespace.  FindFirst returns -1 if the item is not found anywhere in the collection.
The full expression for the update data action inside the loop is
AddItem(Task.GoodStatusIndices, Task.index)




------------------------------
Melissa Bailey
Genesys - Employees

Original Message:
Sent: 01-27-2022 10:36
From: Mo Ford
Subject: Count in Reverse Order

Thank you, Melissa. The second suggestion is much cleaner and worked for finding the count.

If I wanted to instead find the location of the items in the string below that are not EXPIRED, CLOSED or REJECTED with the intent to save the location to be used later in a GetAt function, how would I go about doing that?

EXPIRED, ISSUED, CLOSED, OPENED, REJECTED

------------------------------
Mo Ford
Upgrade

Original Message:
Sent: 01-25-2022 09:46
From: Melissa Bailey
Subject: Count in Reverse Order

Use a loop to go through each item in your collection.  In this example Task.fruit is the string collection.  Name the index variable anything you like.  The maximum loop count tells the loop when to stop, in your case you want the number of items in your list, so use the expression Count(Task.fruit) which will be 4.  Normally people want to go through the list from the start, so inside the loop you would use Task.fruit[Task.index].  You know collection indices start at 0 right?  So
0=Apple
1=Pear
2=Banana
3=Banana
In your case you want to go through it backwards, so instead of Task.fruit[Task.index] you want to look at Task.fruit[Count(Task.fruit)-1-Task.index].
When Task.index=0, you will look at the item with index 4-1-0 = 3 and get Banana
When Task.index=1, you will look at the item with index 4-1-1 = 2 and get Banana
When Task.index=2, you will look at the item with index 4-1-2 = 1 and get Pear
When Task.index=3, you will look at the item with index 4-1-3 = 0 and get Apple

FYI there's a much simpler way of counting how many items are not Banana in your collection.  You could just evaluate this expression, no looping necessary.
Count(Task.fruit)-Count(Find(Task.fruit, "Banana"))
Find returns a collection of indices for which the item exists in the given collection (in your case you would get a collection with the values 2 & 3).  The count of that is 2, and subtracting it from the total count for the collection gets 2.

------------------------------
Melissa Bailey
Genesys - Employees

Original Message:
Sent: 01-25-2022 06:00
From: Mo Ford
Subject: Count in Reverse Order

Hello,

Is there a way in Architect to COUNT in reverse? For example, if I have an array of {Apple, Pear, Banana, Banana} and I want to use the COUNT function but start from the right, excluding anything Banana, how would I go about that? I considered using the REVERSE function but do not want to reverse the actual words, just the order I'm counting them in.

Regards,
Mo
#ArchitectureandDesign
#Routing(ACD/IVR)

------------------------------
Mo Ford
Upgrade
------------------------------

I would think you would want to create another collection of "dateTimeDiffs" between the current time and the dates in your existing collection. The output of DateTimeDiff isn't always friendly (sometimes you get something like this: PT-4337H-59M), so you would need to convert it to a standard time unit, such as seconds.  The most recent date would be the lowest seconds value of course. Then, you would need to run through all of the dateTimeDiffs and use a decision step to see if the current step is less than the previous step (you'd want to start out with a comparator of a really high number so that the you properly loop through all of the items in the array. You then replace the <most recent date placeholder > value if it is lower.

------------------------------
Peter Stoltenberg
Avtex Solutions, LLC
------------------------------

Original Message:
Sent: 02-22-2022 15:18
From: Mo Ford
Subject: Count in Reverse Order

Thanks for all of your help @Melissa Bailey! Last question (for now), if the collection is a collection of dates, is there a way to identify the most recent date? I assume some form of the date that's closest to today.​

------------------------------
Mo Ford
Upgrade

Original Message:
Sent: 02-17-2022 10:42
From: Melissa Bailey
Subject: Count in Reverse Order

You mean get the last item?

GetAt(Task.GoodStatusIndices, Count(Task.GoodStatusIndices)-1)​

------------------------------
Melissa Bailey
Genesys - Employees

Original Message:
Sent: 02-15-2022 04:54
From: Mo Ford
Subject: Count in Reverse Order

Thanks, Melissa. That suggestion worked to gather the locations of the good statuses. If I wanted to reference the highest number in the good status indices, how would I go about doing that in a GetAt() function?

------------------------------
Mo Ford
Upgrade

Original Message:
Sent: 01-31-2022 13:10
From: Melissa Bailey
Subject: Count in Reverse Order

First create a string collection Task.statuses from your string using the Split function.  Create an integer collection Task.GoodStatusIndices to hold the indices (locations) of the items that don't match.  Create another string collection Task.BadStatuses with EXPIRED, CLOSED, & REJECTED as the items.  Then use a loop to go through each item in Task.statuses and determine if it is not in Task.BadStatuses.  If so, add the index to Task.GoodStatusIndices.

The full expression in the decision action is
FindFirst(Task.BadStatuses, Trim(Task.statuses[Task.index])) == -1
Trim is getting rid of any leading or trailing whitespace.  FindFirst returns -1 if the item is not found anywhere in the collection.
The full expression for the update data action inside the loop is
AddItem(Task.GoodStatusIndices, Task.index)




------------------------------
Melissa Bailey
Genesys - Employees

Original Message:
Sent: 01-27-2022 10:36
From: Mo Ford
Subject: Count in Reverse Order

Thank you, Melissa. The second suggestion is much cleaner and worked for finding the count.

If I wanted to instead find the location of the items in the string below that are not EXPIRED, CLOSED or REJECTED with the intent to save the location to be used later in a GetAt function, how would I go about doing that?

EXPIRED, ISSUED, CLOSED, OPENED, REJECTED

------------------------------
Mo Ford
Upgrade

Original Message:
Sent: 01-25-2022 09:46
From: Melissa Bailey
Subject: Count in Reverse Order

Use a loop to go through each item in your collection.  In this example Task.fruit is the string collection.  Name the index variable anything you like.  The maximum loop count tells the loop when to stop, in your case you want the number of items in your list, so use the expression Count(Task.fruit) which will be 4.  Normally people want to go through the list from the start, so inside the loop you would use Task.fruit[Task.index].  You know collection indices start at 0 right?  So
0=Apple
1=Pear
2=Banana
3=Banana
In your case you want to go through it backwards, so instead of Task.fruit[Task.index] you want to look at Task.fruit[Count(Task.fruit)-1-Task.index].
When Task.index=0, you will look at the item with index 4-1-0 = 3 and get Banana
When Task.index=1, you will look at the item with index 4-1-1 = 2 and get Banana
When Task.index=2, you will look at the item with index 4-1-2 = 1 and get Pear
When Task.index=3, you will look at the item with index 4-1-3 = 0 and get Apple

FYI there's a much simpler way of counting how many items are not Banana in your collection.  You could just evaluate this expression, no looping necessary.
Count(Task.fruit)-Count(Find(Task.fruit, "Banana"))
Find returns a collection of indices for which the item exists in the given collection (in your case you would get a collection with the values 2 & 3).  The count of that is 2, and subtracting it from the total count for the collection gets 2.

------------------------------
Melissa Bailey
Genesys - Employees

Original Message:
Sent: 01-25-2022 06:00
From: Mo Ford
Subject: Count in Reverse Order

Hello,

Is there a way in Architect to COUNT in reverse? For example, if I have an array of {Apple, Pear, Banana, Banana} and I want to use the COUNT function but start from the right, excluding anything Banana, how would I go about that? I considered using the REVERSE function but do not want to reverse the actual words, just the order I'm counting them in.

Regards,
Mo
#ArchitectureandDesign
#Routing(ACD/IVR)

------------------------------
Mo Ford
Upgrade
------------------------------