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I am working with SP2019 and On the file share, there is a Legal folder added which gets lot of files dumped into it. Once users work on files, they save it there and send it to external party for review, this file as an example be named as ‘General Accounting Assessment’ and post review it comes back as ‘General Accounting Assessment_V1’ and sits on the same shared folder as the original file.
This review process happen multiple times and that results into 4-5 files on the file share with below names: 1. ‘General Accounting Assessment’ 2. ‘General Accounting Assessment_v1’ 3. ‘General Accounting Assessment_v3’ 4. ‘General Accounting Assessment_v4’
Users initially save file ‘General Accounting Assessment’to SharePoint prior to sending out for review to external party so when v1, v2,v3 and v4 version file arrives, they manually rename the files to match it to name ‘General Accounting Assessment’ and drop into SharePoint so it can add up as a new version to the existing file.
I would like to automate this process and for which I am wondering if there is a way maybe by Flow or SharePoint to check whenever a new file is added, strip off the portion _v1 or _v2 or _v3 or _v4 dynamically appended at the end to file name ‘General Accounting Assessment’ and save it to SharePoint
Can someone please help on the proposed solution, thanks in advance.
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In the basic rules the The Adventuring Day is described as follows:
Assuming typical adventuring conditions and average luck, most adventuring parties can handle about six to eight medium or hard encounters in a day. If the adventure has more easy encounters, the adventurers can get through more. If it has more deadly encounters, they can handle fewer.
In reality many home games, mine included, often don’t have as many as six to eight encounters, covering a range of difficulties, between long rests.
The ‘five minute adventuring day’ takes this to the extreme and describes a situation where PCs are allowed to long rest between pretty much every encounter, and thus don’t have to worry about the ongoing effects of resource depletion from one encounter to another.
It is fairly well established that this is bad for game balance and makes the PCs much stronger (see Are there rules for shortening the adventuring day and reducing the number encounters without unbalancing them? amongst others). However, this questions is not about game balance (i.e. PCs v DM game balance).
I’d like to know how the five minute adventuring day affects intra-party class balance specifically.
- It’s well established that Paladins have extremely powerful burst damage but which other classes benefit most from a shorter adventuring day?
- Which classes are most adversely affected?
- Are all non-spellcasting classes going to be outdamaged by all spellcasting classes under these conditions?
- Does the answer to this question vary much from low to high level play?
For the purposes of this question please assume two hard / deadly encounters, lasting three – five rounds each, and no short rests in an adventuring day – a not uncommon occurrance in my home games. Consider basic class features but don’t worry about archetypes.
I saw this Puzzling problem and thought I would try to write a Python program to solve it. The task is to transform “four” to “five”, forming a new four-letter word at each step, replacing one letter at each step, in as few steps as possible.
But turns out I don’t know how to optimize recursion, so I’m posting here for help. I’m mostly just confused on why the code to change the
past needs to be at the top of the function, but I would also like advice on how to speed this up in general. Right now it takes about 10x as long for each step up
max_depth gets on my computer.
There won’t be any matches until you change
max_depth – I didn’t want anyone copy-pasting and lagging out. There should be a solution at depth 5, according to Puzzling. However, my
words file doesn’t have the
Foud or the word
Fous, which that answer uses. Bumping up to
max_depth six will take my computer ~10 minutes, which I don’t want to try yet.
def hamming(string1, string2): assert len(string1) == len(string2) return sum(char1 != char2 for char1, char2 in zip(string1, string2)) max_depth = 3 start_word = "five" end_word = "four" all_words = open("/usr/share/dict/words", "r").read().lower().splitlines() all_words = list(filter(lambda word: word.isalpha(), all_words)) all_words = list(filter(lambda word: len(word) == len(start_word), all_words)) sequences =  def search(current_word, past = ): # Needs to be first to be fast for some reason past = past[:] past.append(current_word) if len(past) > max_depth: sequences.append(past) return for word in all_words: if hamming(word, current_word) == 1 and word not in past: search(word, past) search(start_word) sequences = [sequence[:sequence.index(end_word) + 1] for sequence in sequences if end_word in sequence] if len(sequences) == 0: print("No matches") else: print(min(sequences, key=len))
I was given a task to find the value of variable a,b,c,d,e and f. But I’m not sure it is even possible, given that only 5 equations are available. Can anybody point out how to solve these:
a+b+c=164.35; d+e+f=94.44; a^2+d^2=20.06^2; b^2+e^2=74.34^2; c^2+f^2=123.27^2
I cannot understand how to define those fields… What are the fields, can anyone plz provide five of those fields..
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- How should a family with young children apply for UK visas? 1 answer
How can I apply for a visit visa to the UK for a family of five – myself, my wife, my son, my daughter-in-law and my granddaughter? Can I apply for all visas under one ID? If yes, how?