I have a bit of a dilemma with the IF COUNT FIND functions in Excel. I am currently researching goal times in soccer matches and have various games with goal times shown as follows (as an example);
23;56;85 (this would be in cell C2 for purpose of this question) The semi colons cannot be changed, this is how they are shown in the data I have, over 50,000 rows of data.
If I apply the formula below into a cell it will give the value "1" as it recognises the number 56.
=IF(COUNT(FIND({56},C2))>0, 1, 0)
However the formula does not work when I ask it to search for any single digit number from 1 to 9, for example the formula below would return the value "1" also because it is picking up the number 5 from the times of 56 and 85.
=IF(COUNT(FIND({5},C2))>0, 1, 0)
How can I adjust to formula so it picks up single digit goal times only? Any helps appreciated!
You can try to match the surrounding semi-columns in the process:
=IF(COUNT(FIND({";56;"},";"&C2&";"))>0, 1, 0)
Notice that (in both formulas) we match something surrounded by ;, but we also surrounded the searched string with ; at each side to handle the edges positions.
Your formula can be simplified further:
=--ISNUMBER(FIND(";56;",";"&C2&";"))
But your initial formula is good if you want later to match one of many values, which you can put in the array.
Related
My worksheet contains orders from clients. The orders are all wooden panels.
Every order is assigned a number which is led by the letter Q.
Column B contains the number of parts in the order.
Column C contains the total m² in the order.
Orders that contain one or more parts that are 2.8 x 0.0735 m will get a row of their own.
I'm trying to count the number of times that this part occurs in a list of more than a thousand rows.
So if I divide the total m² by the m² of the part I'm looking for and divide this by the amount of parts in the order, I should get exactly 1 as a result.
If I take the sum of all the number of parts that result in a 1, I get my total.
Now I'd like to put this in one formula for the entire worksheet, but SUMIF doesn't work the way I'm trying. (It's in Dutch)
=SOM.ALS(B:B;(C:C/(2,8*0,0735)/B:B)=1)
I can't seem to use this formula as a criterium in the SUMIF.
For now I use a helping column that gives the right amount per row. Then take the total SUM of these.
Is it possible to put this in a single formula?
Yes, it is possible. Try this one:
{=SUM(--(B:B=C:C/(2.8*0.0735))*IF(ISERROR(1/B:B),0,1))}
Remember to enter it as an array function with CNTRL + SHIFT + ENTER.
The first half of the formula is just a logical test, after the asterisk it tests if 1/B results in an error (thereby omitting text, zeroes, and blanks) and returns a zero if there is an error.
These are then summed and the result displayed.
In Dutch and English:
{=SOMPRODUCT(--(B:B=(ALS(ISTEKST(C:C);1;C:C))/(2,8*0,0735));B:B)}
{=SUMPRODUCT(--(B:B=(IF(ISTEXT(C:C),1,C:C))/(2.8*0.0735)),B:B)}
is working perfectly. (Enter with Ctrl-Shift-Enter)
The first bit is the logical test, which will check if B:B = C:C / (2.8*0.0735)
It got stuck on #VALUE! because there is text in C:C.
The IF(ISTEXT)) eliminates text by converting them to numeric values, in this case 1, but it can be any numeric value.
The logical test will return TRUE(1) or FALSE(0) because of the double dash or unary operator and this will be multiplied by their respective B:B value.
Because the row with text has no value in B:B, it will result as zero.
I got a problem with my Excel, where it shows a same name twice, if the 'racers' have the excact same time. For example in the picture the racers 7 & 9 and racers 5 & 10 have the same time, but in the Start grid it shows the same name twice. It should be 4. Racer7 5. Racer9 & 9. Racer5 10. Racer10
The Function of Cell I3 =IF(OR(ISBLANK(B3);ISBLANK(C3));"";INDEX($B$3:$B$32;MATCH(J3;$D$3:$D$32;0))) (I have to use format ';' instead of ',') Function of cell J3 =IFERROR(SMALL($D$3:$D$32;H3);"")
Link to the file (does not work in Google Sheets & the functions have to use local formatting)
xlsx file
MATCH and RANK Getting Sick When Handling Time
Formulas
COMMA
[D3] =IF(NOT(ISNUMBER(C3)),"",ROUND($D$1-A3*"00:10,0"-C3,8))
[H3] =IF(ISNUMBER(K3),RANK(J3,D$3:D$32,2),"")
[I3] =IF(IFERROR(INDEX($B$3:$B$32,SMALL(IF($D$3:$D$32=J3,ROW($J$3:$J$32)-ROW(J$3)+1),COUNTIF($J$3:$J3,J3))),"")=0,"",IFERROR(INDEX($B$3:$B$32,SMALL(IF($D$3:$D$32=J3,ROW($J$3:$J$32)-ROW(J$3)+1),COUNTIF($J$3:$J3,J3))),""))
[J3] =IFERROR(SMALL($D$3:$D$32,A3),"")
[K3] =IFERROR(J3-J$3,IF(I3="","","disqualified"))
COLON
[D3] =IF(NOT(ISNUMBER(C3));"";ROUND($D$1-A3*"00:10;0"-C3;8))
[H3] =IF(ISNUMBER(K3);RANK(J3;D$3:D$32;2);"")
[I3] =IF(IFERROR(INDEX($B$3:$B$32;SMALL(IF($D$3:$D$32=J3;ROW($J$3:$J$32)-ROW(J$3)+1);COUNTIF($J$3:$J3;J3)));"")=0;"";IFERROR(INDEX($B$3:$B$32;SMALL(IF($D$3:$D$32=J3;ROW($J$3:$J$32)-ROW(J$3)+1);COUNTIF($J$3:$J3;J3)));""))
[J3] =IFERROR(SMALL($D$3:$D$32;A3);"")
[K3] =IFERROR(J3-J$3;IF(I3="";"";"disqualified"))
Why is MATCH 'miscalculating' to '7' instead of '6' in cells 'I6' and 'I7' in OP's worksheet (formula in 'D3')?
Time has a ton of decimals so I guess it's 'seeing' the values in 'D8' and 'D9' as different values. To avoid this you can round the values. If you want to use only these values it is enough to round them to 8 decimals for the numbers to be recognized as different even by a millisecond. If you want to sum them there might be some inaccuracies. In OP's case 8 decimals is more than enough.
RANK (formula in 'H3') is also 'miscalculating' if no rounding.
Why the long formula?
Best try it with and without the IF statement and see for yourself.
Here's a Hint:
For this you need a tie breaker. The unfortunately best way of doing this is using a helper column. In my test sheet I used column E but the column could, of course, be anywhere. More importantly, it could be hidden. In this column you enter a formula like
=D3+ROW()/10^8
The point is that the addition must be so small that it makes no difference to the result on rounding. So, the number of results you treat in this may makes a difference. If you find that the addition changes the result in the last row, increase the exponent. The change I made add 0.0001 seconds to each result, multiplied by the row number: 0.0001 in the first row, 0.0002 in the second, 0.0003 in the third etc. Check the results in the 10th and 100th row.
Now the results in column E are all different and it's these results that are used in columns J and I.
[J3] =SMALL($E$3:$E$32,H3)
and
[I3] =INDEX($B$3:$B$32,MATCH(J3,$E$3:$E$32,0))
There will be no more duplicates but the "winner" of a draw is decided by his position in the list.
A solution in old school array-formula-style would be:
Note It's an array-formula which needs to be confirmed through CTRLSHIFTENTER
=IF(OR(ISBLANK(B3),ISBLANK(C3)),"",INDEX($B$1:$B$32,SMALL(IF($D$1:$D$12=J3,ROW($J$1:$J$12)),COUNTIF($J$1:$J3,J3))))
The IF checks the timelist for the current time value and gives all matching lines back which are getting ranked by small. COUNTIF counts the occurences of the current time up to the current line.
I have personal ID's in reports I have to find in one cell. Too bad the string in the cell which hides this ID can be anything, the ID can be at the beginning, the end, anywhere, but it is there.
The only thing I know is the pattern "space,letter,letter,number,number,number,number,number,number,space". Jike DB544345
I was looking for the correct word for this "mask", but couldn't find an answer. Thank you for your help.
As the comments are numerous I have created a minimal example that might represent what the OP is dealing with:
A1: 123456789 DB544345 asdfg asdfghjk
A2: creating dummy data is a DB544345 pain
A3: DB5443456 and soething else
parsed a copy of that in ColumnB with Text To Columns (with space as the delimiter) then applied:
=IFERROR(IF(AND(LEN(B1)=8,CODE(LEFT(B1))>64,CODE(LEFT(B1))<91,CODE(MID(B1,2,1))>64,CODE(MID(B1,2,1))<91,ISNUMBER(RIGHT(B1,6)*1),RIGHT(B1,6)*1>99999),B1,""),"")
to K1, copied this across to P1 and then K1:P1 down.
A concise "built-in function only" solution to a problem such as this requires a bit of tinkering as many attempts will dead-end or need workarounds due to deficiencies and quirks in the built-in Excel formulas. I much prefer single cell formulas because they minimally affect the general spreadsheet structure. However, due to the limitations listed above, complex single cell solutions often come at the cost of being rather long and cumbersome (this answer is somehow still only two lines on my formula bar in Excel). I came back to your question and cobbled together a formula that can (as far as I have tested) extract the first occurrence of this pattern with a single cell formula. This is an array formula (Ctrl+Shift+Enter instead of Enter) that assumes your data is in A2. This rough formula returns the first 8 characters if no match is found and throws #REF if the string is shorter than 10 characters.
=MID(A2,MIN(IF(MID(A2,ROW(INDIRECT("A1:A"&(LEN(A2)-9))),1)=" ",IF(MID(A2,ROW(INDIRECT("A1:A"&(LEN(A2)-9)))+9,1)=" ",IF(CODE(MID(A2,ROW(INDIRECT("A1:A"&(LEN(A2)-9)))+1,1))>64,IF(CODE(MID(A2,ROW(INDIRECT("A1:A"&(LEN(A2)-9)))+1,1))<91,IF(CODE(MID(A2,ROW(INDIRECT("A1:A"&(LEN(A2)-9)))+2,1))>64,IF(CODE(MID(A2,ROW(INDIRECT("A1:A"&(LEN(A2)-9)))+2,1))<91,IF(IFERROR(MID(A2,ROW(INDIRECT("A1:A"&(LEN(A2)-9)))+3,6)*1>99999,FALSE),ROW(INDIRECT("A1:A"&(LEN(A2)-9)))))))))))+1,8)
Let me try to break this down at least on a high level. We are splitting the main text into every possible ten character chunk so that we can test each one using the suggestion of #pnuts to verify the Unicode values of the first two characters and run an ISNUMBER check on the rest of the string. This first block recurs throughout my formula. It generates a list of numbers from 1 to n-9 where n is the length of our main text string.
ROW(INDIRECT("A1:A"&(LEN(A2)-9)))
Let's assume our string is 40 characters long and replace the above formula with {1...31}. Using this number sequence generation we can check if characters 1 to 31 are spaces:
IF(MID(A2,{1...31},1)=" "
Then we can check if characters 10 to 40 are spaces:
IF(MID(A2,{1...31}+9,1)=" "
Then we can check if characters 2 to 32 are capital letters:
IF(CODE(MID(A2,ROW(INDIRECT("A1:A"&(LEN(A2)-9)))+1,1))>64,
IF(CODE(MID(A2,ROW(INDIRECT("A1:A"&(LEN(A2)-9)))+1,1))<91
Then we can check if characters 3 to 33 are capital letters:
IF(CODE(MID(A2,ROW(INDIRECT("A1:A"&(LEN(A2)-9)))+2,1))>64,
IF(CODE(MID(A2,ROW(INDIRECT("A1:A"&(LEN(A2)-9)))+2,1))<91
Then we can check if the strings of characters 4 to 9, 5 to 10, ..., 33 to 38, 34 to 39 are six-digit numbers:
IF(IFERROR(MID(A2,ROW(INDIRECT("A1:A"&(LEN(A2)-9)))+3,6)*1>99999,FALSE)
If all conditions are TRUE, that 10 digit chunk test will return the index of its first character in the string via another instance of the original array {1...31}. Otherwise it returns nothing. We take the Min of all return indexes and then use the Mid function to grab the 8 digit string determined by the aforementioned minimum index:
=MID(A2,MIN(matching index list)+1,8)
I think this will work, if we assume that the SPACE at the beginning and end are merely to differentiate the ID from the rest of the string; hence would not be present if the ID were at the beginning or end of the string. This formula is case insensitive. If case sensitivity is required, we could do character code comparisons.
=LOOKUP(2,1/((LEFT(myArr,2)>="AA")*(LEFT(myArr,2)<="ZZ")*(LEN(myArr)=8)*ISNUMBER(-RIGHT(myArr,6))),myArr)
Where myArr refers to:
=TRIM(MID(SUBSTITUTE(TRIM(Sheet2!A1)," ",REPT(" ",99)),(ROW(INDIRECT("1:10"))-1)*99+1,99))
If myArr is initially defined with the cursor in B1, referring to A1 as shown, it will adjust to refer to the cell one column to the left of the cell in which the Name appears.
The 10 in 1:10 is the maximum number of words in the string -- can be adjusted if required.
I have variating numeric entries (SF123456, SF142365, ...). Every number of the numeric entries corresponds to a specific code. For each number of each entry I need to enter on a separate cell the corresponding code (download here example sheet: www.nivpat.com/Example.zip) How can I create an automatic function as I have thousands of entries to divide into codes... thanks!
Alright. What I did to solve this one is this:
Remove the '=' sign in your match table to be able to do a VLOOKUP on it;
Add the position of the digit you want to look up in the row 9 right above the headings. You might want to hide this row for cleaner presentation;
I used the following formula in the cells to extract the values:
=VLOOKUP(VALUE(MID($A11, B$9, 1)), $A$2:$B$7, 2, 0)
The VLOOKUP does the lookup on your table in A2:B7. The MID() extract exactly one character beginning with the character specified in B9 (in this case it would be 3). And the VALUE() converts the text string to a number to be able to do a match with the table above.
The only thing you now have to do is to drag your formulas and it's working !
I am have a string with 6 spaces, e.g. 000000. Each space can hold one of three digits - 0, 1, or 2. I know that I can get a total of 120 permutations using the Permut function in Excel, i.e. =PERMUT(6,3) = 120. But I would actually like to have each individual permutation in a cell, e.g. 000001, 000010, etc.. Ideally, the end result would be 120 rows of unique 6-digit IDs.
Please help if you know a faster way of accomplishing this without entering the figures manually.
Thanks!
There is a VBA functionin the last post on this page. Copy it into a VBA module, then in Excel, create a column of integers from 0 to n where n = the number of IDs you want. In the next column, call the VBA function with the value from the first column as the first argument, and 3 as the second argument. Something like
Column A Column b
0 =baseconv(A1, 3)
1 =baseconv(A2, 3)
2 =baseconv(A3, 3)
... etc.
Your IDs are really just incremental values using a base 3 counting system. You can format the output to get leading zeros with a custom format of '000000'.
Incidentally, with 6 positions and 3 available values, you can get 3 ^ 6, or 729 unique IDs
First, I don't think you're using PERMUT correctly here. What PERMUT(6,3) gives you is the total number of ways to arrange three things picked out of a set of six things. So the result is 120 because you could have 6*5*4 possible permutations. In your case you have 3^6 = 729 possible strings, because each position has one of three possible characters.
Others have posted perfectly fine VBA-based solutions, but this isn't that hard to do in the worksheet. Here is an array formula that will return an array of the last six digits of the ternary (base-3) representation of a number:
=FLOOR(MOD(<the number>,3^({5,4,3,2,1,0}+1))/(3^{5,4,3,2,1,0}),1)
(As WarrenG points out, just getting a bunch of base-3 numbers is one way to solve your problem.)
You would drag out the numbers 0 through 728 in a column somewhere, say $A$1:$A$729. Then in $B$1:$G$1, put the formula:
=FLOOR(MOD(A1,3^({5,4,3,2,1,0}+1))/(3^{5,4,3,2,1,0}),1)
remembering to enter it as an array formula with Ctrl-Shift-Enter. Then drag that down through $B$729:$G$729.
Finally in cell $H$1, put the formula:
=CONCATENATE(B1,C1,D1,E1,F1,G1)
and drag that down through $H$729. You're done!