I need to generate a list that contains all possible 4 character strings made up of the characters A,B,C,D,E,F. What is the best way to accomplish this? Characters can be used any number of times.
list("".join(l) for l in itertools.product('ABCDEF', repeat=4))
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What im working with
I have a list of product names, but unfortunately they are written in uppercase I now want to make only the first letter uppercase and the rest lowercase but I also want all words with 3 or less symbols to stay uppercase
im trying if functions but nothing is really working
i use the german excel version but i would be happy if someone has any idea on how to do it im trying different functions for hours but nothing is working
=IF(LENGTH(C6)<=3,UPPER(C6),UPPER(LEFT(C6,1))&LOWER(RIGHT(C6,LENGTH(C6)-1)))
but its a #NAME error excel does not recognize the first and the last bracket
This is hard! Let me explain:
I do believe there are German words in the mix that are below 4 characters in length that you should exclude. My German isn't great but there would probably be a huge deal of words below 4 characters;
There seems to be substrings that are 3+ characters in length but should probably stay uppercase, e.g. '550E/ER';
There seem to be quite a bunch of characters that could be used as delimiters to split the input into 'words'. It's hard to catch any of them without a full list;
Possible other reasons;
With the above in mind I think it's safe to say that we can try to accomplish something that you want as best as we can. Therefor I'd suggest
To split on multiple characters;
Exclude certain words from being uppercase when length < 3;
Include certain words to be uppercase when length > 3 and digits are present;
Assume 1st character could be made uppercase in any input;
For example:
Formula in B1:
=MAP(A1:A5,LAMBDA(v,LET(x,TEXTSPLIT(v,{"-","/"," ","."},,1),y,TEXTSPLIT(v,x,,1),z,TEXTJOIN(y,,MAP(x,LAMBDA(w,IF(SUM(--(w={"zu","ein","für","aus"})),LOWER(w),IF((LEN(w)<4)+SUM(IFERROR(FIND(SEQUENCE(10,,0),w),)),UPPER(w),LOWER(w)))))),UPPER(LEFT(z))&MID(z,2,LEN(v)))))
You can see how difficult it is to capture each and every possibility;
The minute you exclude a few words, another will pop-up (the 'x' between numbers for example. Which should stay upper/lower-case depending on the context it is found in);
The second you include words containing digits, you notice that some should be excluded ('00SICHERUNGS....');
If the 1st character would be a digit, the whole above solution would not change 1st alpha-char in upper;
Maybe some characters shouldn't be used as delimiters based on context? Think about hypenated words;
Possible other reasons.
Point is, this is not just hard, it's extremely hard if not impossible to do on the type of data you are currently working with! Even if one is proficient with writing a regular expression (chuck in all (non-available to Excel) tokens, quantifiers and methods if you like), I'd doubt all edge-case could be covered.
Because you are dealing with any number of words in a cell you'll need to get crafty with this one. Thankfully there is TEXTSPLIT() and TEXTJOIN() that can make short work of splitting the text into words, where we can then test the length, change the capitalization, and then join them back together all in one formula:
=TEXTJOIN(" ", TRUE, IF(LEN(TEXTSPLIT(C6," "))<=3,UPPER(TEXTSPLIT(C6," ")),PROPER(TEXTSPLIT(C6," "))))
Also used PROPER() formula as well, which only capitalizes the first character of a word.
I'm in Dyalog 17 and would like to generate unique names to be used with its graphics object library. So, for example, I have the letter 'l' and want to take the number 1, convert it to a character and then concatenate the two together to form 'l1'. This is such trivial stuff in other languages but I can't find the documentation explaining how to do this in APL. Thanks for your help!
To concatenate the letter 'l' to the number 1 to form the characters 'l1' you do this:
'l',⍕1
The system function ⎕FMT can be of use here. For example:
'P<I>ZI7' ⎕FMT ⍳10
I000000
I000001
I000002
I000003
I000004
I000005
I000006
I000007
I000008
I000009
The format string specifies to format the numbers as integers, in width of 7, zero filling, with a positive left decoration of the letter'I'.
I'm on APL2 in the Mainframe, so my answer might not be exactly what you're after, but here's how I would do it:
∊⍕¨'L',1
So first catanate the letter and the numeric digits. Then FORMAT EACH to produce a vector of character scalars. Finally, ENLIST to produce a simple vector.
This is a slight generalization of SteveH's reply. More general in the sense that it handles input strings (rather than scalars) and works equally well regardless if the digit or letter comes first.
I have the following string :
var myStr = "abc12ef4567gh90ijkl789"
The size of the list is not fixed and it contains number in between. I want to extract the numbers and store them in the form of a list in this manner:
List(12,4567,90,789)
I tried the solution mentioned here but cannot extend it to my case. I just want to know if there is any faster or efficient solution instead of just traversing the string and extracting the numbers one by one using brute force ? Also, the string can be arbitrary length.
It seems you may just collect the numbers using
("""\d+""".r findAllIn myStr).toList
See the Scala demo. \d+ matches one or more digits, findAllIn searches for multiple occurrences of the pattern inside a string (and also un-anchors the pattern so that partial matches could be found).
If you prefer a splitting approach, you might use
myStr.split("\\D+").filter(_.nonEmpty).toList
See another demo. Here, \D+ matches one or more non-digit chars, and these chunks are used to split on (texts between these chunks land in the result). .filter(_.nonEmpty) will remove empty items that usually appear due to matches at the start/end of the string.
While there are a lot of solutions for how to find all the (unique) permutations of a string of unique characters, I haven't found solutions that work when the characters are non-unique. I have listed out my idea below and would appreciate feedback, but also feel free to provide your own ideas.
My idea:
To illustrate my algorithm, I'm using the example of the string ABBC, which I want to find all permutations of. Since there are two B's I will be labelling them B1 and B2.
Create a new string by removing all duplicate characters from the original string (e.g. turn AB1B2C into AB1C).
Find all possible permutations of the new string (e.g. AB1C, ACB1, B1AC, etc.). There are many algorithms to do this, since the string's characters are all unique.
Choose one duplicate character. For each permutation, insert the chosen duplicate characters at every "position" of the permutation, except when the character just before the duplicate character has the same value as the duplicate character (e.g. For the permutation AB1C, since the duplicate character is B2, insert it to get B2AB1C, AB2B1C, AB1CB2. Exception: Don't do AB1B2C, since that's just a duplicate of AB2B1C).
Continue to do step 3 but now choose a different duplicate character. (Do this until all duplicate characters have been chosen exactly once.)
Prior research: The answer by Prakhar on this SO question claims to work for duplicates: Generate list of all possible permutations of a string. It might, but I suspect there's a bug in the code.
How about this: suppose that the string with duplicates is of length N. Now consider the sequence 0,1,...N-1. Find all its permutations using one of the known algorithms. For each permutation in this list, generate a corresponding string by using the number in the permutation as an index into the original string. For example, if the string is ABBC, then the sequences will be 0,1,2,3; 0,1,3,2; etc. The sequence 3,0,1,2, as an example, is one of the permutations, and it yields the string CABB
I need to randomly swap out the first letter of a word with one of the other letters. What i am having trouble is with specifying that i only need to randomly generate ONE character. I cant use any conditionals, so can anyone please recommend a method to use?