In my code, legends are running within a loop, and I am trying to show a graph with
0-10%
10-20%
and so on. The problem is when I write this code
legend->AddEntry(gr[i], Form("%d0-%d0 %%",i+0,i+1), "lep");
It shows
00-10%
10-20% etc
So how to not show 00, but 0 in the first line?
A small adaptation of the shown statement should be enough; use:
legend->AddEntry(gr[i], Form("%d-%d %%", i*10 , (i+1)*10), "lep");
Explanation:
Form("%d0-%d0 %%",i+0,i+1) seems to be some kind of string formatting, and i your loop variable which runs from 0 to 9, right? The shown Form statement just appends "0" hard-coded to the single digit in i; instead, you can multiply i by 10, resulting in the actual numbers you want printed; and since 10*0 is still 0, this will be a single digit still; so, replace the previous Form(...) call with Form("%d-%d %%", i*10, (i+1)*10) and you should have the result you want!
In case you're worrying that printing i*10 is "less efficient" than printing i with "0" suffix - don't. The formatting and output of the string is most probably orders of magnitude slower than the multiplication anyway, so any overhead of doing multiple multiplications is negligible.
Related
I am using printf to output contents. Now I see the format specification as "%03.3u" in another person's code, as per my understanding the "03" before the dot already specifies the width of the output as 3 digits, and padding with zeros if there are not 3, while the "3" after the dot also specifies that there should be 3 digits output. Therefore, it seems "03" before the dot and "3" after the dot is duplicated.
I make the following tests:
char l[50];
sprintf(l, "%03.3u", 5);
sprintf(l, "%03u", 5);
sprintf(l, "%.3u", 5);
And confirm the output is always 005. So why someone else should use "%03.3u" instead of "%03u" or "%.3u"?
The output will be the same for the particular values you have used. The number before the . is the minimum field width while the number after (for the u conversion specifier, at least) it is the minimum number of digits to output. You can see the difference between the two with something like:
printf("%3.2u\n", 7)
which gives you space07 - minimum two digits output and minimum three characters wide.
However, the fact that you have the numbers the same means that you'll get three digits minimum in a field at least three characters wide. Even if you had used %03.2u (different minimums), the presence of that 0 means to left-pad with 0 rather than space, so you'd still see 005.
Bottom line is, to get the full three digits, you can use the 0 zero-pad modifier or the minimum digit count modifier but you don't need both.
However, since having both doesn't have any adverse effects beyond forcing people to question the sanity of those that wrote it :-), it's functionally okay.
The 03 is the field width with zero-padding. This means that a minimum of 3 characters are to be output, and if there were fewer than three, left-pad with zeroes.
The second 3 is the minimum number of digits to output.
When both of these are specified, the precision will be applied, and if the result is narrower than the minimum field width, then the output will be padded. For exampleprintf("q%6.3u", 5) will produce q 005 . (I use the q because stackoverflow formatting eats the spaces otherwise).
If you're printing an unsigned integer and you didn't use the sign flag, then the number of digits is the same as the field width (since the only output is digits). %03u, %.3u and %03.3u all have the same effect.
I guess the person wrote %03.3u since they did not properly understand the meaning of these things so they guessed something, it worked, and they decided to not make any further changes.
If you print a sign character then the field width differs from the digit count, e.g. you could experiment with %+3u versus %+.3u. Or if you use %d and print a negative number.
How does this code work?
;''
6666,-2%{2+.2/#*\/10.3??2*+}*
`1000<~\;
It seem to use an array #* and a cycle {/**/}, but what is 6666? what is \/?
The first three characters; ;'', are unneeded for the program to function. They simply discard all input and replace it with an empty string, in case your compiler needs an input necessarily.
6666, prints out an array 6666 elements long, each of which are the numbers 0-6665.
-2% is a mapping function. It reverses the function and deletes every two elements. You now you have an array that is 3333 elements long, and it goes [6665 6663 6661 … 5 3 1]
{foo}* is a folding block call. For every element, do the following to the combination of elements. For example, 5,{+}* would add together the numbers 0-4.
So, let's see what we're doing in this folding block call.
2+ add two to the element.
. duplicate the element.
2/ halve it. Your sub-stack looks like this; (n+2),((n+2)/2)
# pulls the third element to the top.
This is the first function we cannot do, since our original stack is only two tall. We'll get back to this later.
*\/ will be skipped for now, we'll get back to it once we discuss folding more.
10.3?? Duplicate 10, then push a 3. [10 10 3]. ? is exponentiation, so we have [10 1000], then again gives us a 1 with 1000 zeroes afterwards.
2* Multiply it by two. So now we have a 2 with 1000 zeroes after.
+ Adds the rest of our math to 2e(1e3)
So, let's go back to that pesky #.
#*\/ will grab the third element and bring it to the top, then multiply it by the next top element ((n+2)/2), then we divide n by this number.
This is an expansion of the Leibniz Series.
\`1000< turns the int into a string, then throws a decimal after the 3.
~ dumps the string into a number again.
\; deleted the rest of the stack.
To answer your specific questions;
6666 was chosen, since half is 3333 (length of array), and we want more than pi times the number of digits of accuracy we want. We could make it smaller if we wanted, but 6666 is a cute number to use.
\/ Is the "inverse division" pair. Take a, take b, then calculate b/a. This is because the \ changes the order of the top two elements in the array, and / divides them.
So I have some code that does essentially this:
REAL, DIMENSION(31) :: month_data
INTEGER :: no_days
no_days = get_no_days()
month_data = [fill array with some values]
WRITE(1000,*) (month_data(d), d=1,no_days)
So I have an array with values for each month, in a loop I fill the array with a certain number of values based on how many days there are in that month, then write out the results into a file.
It took me quite some time to wrap my head around the whole 'write out an array in one go' aspect of WRITE, but this seems to work.
However this way, it writes out the numbers in the array like this (example for January, so 31 values):
0.00000 10.0000 20.0000 30.0000 40.0000 50.0000 60.0000
70.0000 80.0000 90.0000 100.000 110.000 120.000 130.000
140.000 150.000 160.000 170.000 180.000 190.000 200.000
210.000 220.000 230.000 240.000 250.000 260.000 270.000
280.000 290.000 300.000
So it prefixes a lot of spaces (presumably to make columns line up even when there are larger values in the array), and it wraps lines to make it not exceed a certain width (I think 128 chars? not sure).
I don't really mind the extra spaces (although they inflate my file sizes considerably, so it would be nice to fix that too...) but the breaking-up-lines screws up my other tooling. I've tried reading several Fortran manuals, but while some of the mention 'output formatting', I have yet to find one that mentions newlines or columns.
So, how do I control how arrays are written out when using the syntax above in Fortran?
(also, while we're at it, how do I control the nr of decimal digits? I know these are all integer values so I'd like to leave out any decimals all together, but I can't change the data type to INTEGER in my code because of reasons).
You probably want something similar to
WRITE(1000,'(31(F6.0,1X))') (month_data(d), d=1,no_days)
Explanation:
The use of * as the format specification is called list directed I/O: it is easy to code, but you are giving away all control over the format to the processor. In order to control the format you need to provide explicit formatting, via a label to a FORMAT statement or via a character variable.
Use the F edit descriptor for real variables in decimal form. Their syntax is Fw.d, where w is the width of the field and d is the number of decimal places, including the decimal sign. F6.0 therefore means a field of 6 characters of width with no decimal places.
Spaces can be added with the X control edit descriptor.
Repetitions of edit descriptors can be indicated with the number of repetitions before a symbol.
Groups can be created with (...), and they can be repeated if preceded by a number of repetitions.
No more items are printed beyond the last provided variable, even if the format specifies how to print more items than the ones actually provided - so you can ask for 31 repetitions even if for some months you will only print data for 30 or 28 days.
Besides,
New lines could be added with the / control edit descriptor; e.g., if you wanted to print the data with 10 values per row, you could do
WRITE(1000,'(4(10(F6.0,:,1X),/))') (month_data(d), d=1,no_days)
Note the : control edit descriptor in this second example: it indicates that, if there are no more items to print, nothing else should be printed - not even spaces corresponding to control edit descriptors such as X or /. While it could have been used in the previous example, it is more relevant here, in order to ensure that, if no_days is a multiple of 10, there isn't an empty line after the 3 rows of data.
If you want to completely remove the decimal symbol, you would need to rather print the nearest integers using the nint intrinsic and the Iw (integer) descriptor:
WRITE(1000,'(31(I6,1X))') (nint(month_data(d)), d=1,no_days)
=COUNTIFS(Orders!$T:$T,$B4)
is a code that gives 0 or a +ve result
I use this across 1500 cells which makes the sheet gets filled with 0s
I'd like to remove the Zeros by using the following formula
if(COUNTIFS(Orders!$T:$T,$B3,Orders!$F:$F,""&P$1&"*")=0,
"",
COUNTIFS(Orders!$T:$T,$B3,Orders!$F:$F,""&P$1&"*"))
This calculates every formula twice and increases the calculation time.
How can we do this in 1 formula where if the value is 0 - keep empty - otherwise display the answer
I suggest this cell-function:
=IFERROR(1/(1/COUNTIFS(Orders!$T:$T,$B4)))
EDIT:
I'm not sure what to add as explanation. Basically to replace the result of a complex calculation with blank cells if it results in 0, you can wrap the complex function in
IFERROR(1/(1/ ComplexFunction() ))
It works by twice taking the inverse (1/X) of the result, thus returning the original result in all cases except 0 where a DIV0 error is generated. This error is then caught by IFERROR to result in a blank cell.
The advantage of this method is that it doesn't need to calculate the complex function twice, so can give a significant speed/readability increase, and doesn't fool the output like a custom number format which can be important if this cell is used in further functions.
You only need to set the number format for your range of cells.
Go to the menu Format-->Number-->More Formats-->Custom Number Format...
In the entry area at the top, enter the following: #;-#;""
The "format" of the format string is
(positive value format) ; (negative value format) ; (zero value format)
You can apply colors or commas or anything else. See this link for details
instead of your =COUNTIFS(Orders!$T:$T,$B4) use:
=REGEXREPLACE(""&COUNTIFS(Orders!$T:$T,$B4), "^0$", )
also, to speed up things you should avoid "per row formulae" and use ArrayFormulas
I want to write a program that can give me all 4 letter words (from the dictionary or outside the dictionary). I code in C++. And by far, I've reached nowhere.
I'm simply a beginner in C++, I can apply the logic but I'm not introduced to advanced features in C++. It doesn't matter if it takes a long time for this program to end execution, I just want the solution.
For example:
abcd
king
ngik
cbda
play
lpay
payl
and so on (just a few of the millions of outputs I hope this program to output).
NOTE: The words generated need not make sense and I do not want to discard any combinations, I want it all.
I suggest looping say i from 0 to 26^4 - 1, each time outputting 'A' + i / (26*26*26), 'A' + i / (26*26) % 26, 'A' + i / 26 % 26, and 'A' + i % 26, then a newline.
Make a array that has all the possible letters in it(add numbers and symbols if you want).
Then use four nested for loops that loop a number from 0 to the length of the array.
Lets say that the loop number variables are a,b,c,d.
In the inner loop(the last one) you can output it as array[a] + array[b] + array[c] + array[d]
This gives all the possible combinations where you can add numbers and symbols aswell.
use recursion
explore DFS a 26-ary tree and output the letter every time you go down on the corresponding branch. depth = 4.
P.S. recursive algorithms eat stack memory like hell... so be sure your machine let programs have enough stack mem. You don't wanna run this on a pic micro with only 3-level call stack :-)