Based on Apple's documentation here: https://developer.apple.com/library/mac/documentation/Cocoa/Conceptual/Strings/Articles/FormatStrings.html
It's pretty easy to understand the number to the right of the decimal point is the number of digis will be rounded up...
For example, %1.2f, 123456.123456 will turn out 123456.12 and %1.4f will turn out 123456.1234...
But it looks like the number to the left of decimal does nothing.
I tried changing the number to whatever I can think of, nothing happened.
What does it do?
The number before the decimal point in the format is called the format string's width. That is, if the resultant string would involves less characters than its width, it will be left-padded with blank spaces. You don't see any change because you either aren't using a high enough number (try something ridiculous like 100 or 200), or don't have a means of properly seeing your whitespace.
Related
I have a lot of excel files looking like that:
Example:
My goal is to make it look like that:
To do that, I used very simple excel's function:
=F7&" "&G7&".........cat."&" "&H7&" times "&I7&CHAR(10)&F8&" "&G8&".........cat."&" "&H8&" times "&I8&CHAR(10)
The thing is, the number of dots placed before "cat" is not constant. It depends where the previous sentence ends and my formula doesn't take it into account - it always adds 9 dots, which means I have to add the rest of the dots manually.
Any ideas how to make it work? :D
The REPT function can do this. Use LEN to calculate the length of what you're adding the dots to, then subtract that from the desired width of the result. That will repeat the dot enough times to fill the column. For example, if you want the text with dots to be 40 characters, right padded with .:
=F1&" "&G1&REPT(".",40-LEN(G1))&"cat."&" "&H1&" times "&I1&CHAR(10)&F2&""
=LEFT(A1 & REPT(".",22-LEN(A1))&"cat",25)
22 = fixed width - len("cat"), 25 - fixed width.
edit - i revised because my original answer was not correct but I see Comintern has posted a similar response since.
I am working on some legacy code at the moment and have come across the following:
FooString = String.Format("{0:####0.000000}", FooDouble)
My question is, is the format string here, ####0.000000 any different from simply 0.000000?
I'm trying to generalize the return type of the function that sets FooDouble and so checking to make sure I don't break existing functionality hence trying to work out what the # add to it here.
I've run a couple tests in a toy program and couldn't see how the result was any different but maybe there's something I'm missing?
From MSDN
The "#" custom format specifier serves as a digit-placeholder symbol.
If the value that is being formatted has a digit in the position where
the "#" symbol appears in the format string, that digit is copied to
the result string. Otherwise, nothing is stored in that position in
the result string.
Note that this specifier never displays a zero that
is not a significant digit, even if zero is the only digit in the
string. It will display zero only if it is a significant digit in the
number that is being displayed.
Because you use one 0 before decimal separator 0.0 - both formats should return same result.
I've got a column called Amount, with a lot of numbers looking like this:
67000.00000000000000000000
Some of the columns have 2 numbers after the decimal that need to be retained.
Which should amount to $67,000.00
But my problem is, when I format it into currency or numbers, I get MUCH larger numbers than i would like, looking like this:
6.700.000.000.000.000.000.000.000,00
How can I get it into the right format?
Edit: For this scenario, the user was using ACC2013 and the Field Type was Short Text. The method of conversion that succeeded was : CCur(Val(FieldNameHere))
CCur(YourFieldName)
This will convert it to a currency format.
CLng(YourFieldName)
This will convert it to a long integer format. (It will cut off the decimals)
If you're looking for a reference, Microsoft has a few examples and goes into brief detail about some of these conversion functions.
CCur(Replace("67000.00000000000000000000", ".", Format(0, ".")))
You have to replace point symbol to actual decimal separator before conversion. Because you can't know actual seprator, choosen in regional settings, you have to find it out - and such Format() operation does dirty work.
I have some decimal numbers that I need to write to a text file with leading zeros when appropriate. I've done some research on this, and everything I've seen suggests something like:
REAL VALUE
INTEGER IVALUE
IF (VALUE.LT.0) THEN
IVALUE = CEILING(VALUE)
ELSE
IVALUE = FLOOR(VALUE)
ENDIF
WRITE(*,1) IVALUE, ABS(VALUE)-ABS(IVALUE)
1 FORMAT(I3.3,F5.4)
As I understand it, the IF block and ABS parts should allow this to work for all values on -100 < VALUE < 1000. If I set VALUE = 12.3456, the code above should produce "012.3456" as the output, and it does. However if I have something like VALUE = -12.3456, I'm getting "(3 asterisks).3456" as my output. I know the asterisks usually shows up when there are not enough characters provided for in the FORMAT statement, but 3 should be enough in this example (1 character for the "-" and two characters for "12"). I haven't tested this yet with something like VALUE = -9.876, but I'd expect the output to be "-09.8760".
Is there something wrong in my understanding of how this works? Or is there some other limitation of this technique that I'm violating?
UPDATE: Okay I've looked into this some more, and it seems to be a combination of a negative value and the I3.3 format. If VALUE is positive and I have the I3.3, it will put leading zeros as expected. If VALUE is negative and I only have I3 as my format, I get the correct value output, but it will be padded with spaces before the negative sign instead of padded with zeros after the negative (so -9.8765 is output as " -9.8765", but that leading space breaks what I'm using the .txt file for, so it's not acceptable).
Tho problem is with your integer data edit descriptor. With I3.3 you require at least 3 digits and the field width is only 3. There is no place for the minus sign. Use I4.3 or, In Fortran 95 and above, I0.3.
Answer to your edit: Use I0.3, it uses the minimum number of characters necessary.
But finally, you just probably want this: WRITE(*,'(f0.3)') VALUE
Of course, I could get what I'm looking for by changing it up a little bit to
REAL VALUE
INTEGER IVALUE
IF (VALUE.LT.0) THEN
WRITE(*,1) FLOOR(ABS(IVALUE)), ABS(VALUE)-FLOOR(ABS(VALUE))
1 FORMAT('-',I2.2,F5.4)
ELSE
WRITE(*,2) FLOOR(VALUE), ABS(VALUE)-FLOOR(BS(VALUE))
2 FORMAT(I3.3,F5.4)
ENDIF
But this feels a lot clunkier, and in reality I'm going to try to be writing multiple values in the same line, which will lead to really messy IF blocks or complex cursor movement, which I'd like to avoid if at all possible.
as another way to skin the cat.. I'd prefer not to do arithmatic on the data at all but just work on the format:
character*8 fstring/'(f000.4)'/
val=12.34
if(val.gt.1)then
write(fstring(3:5),'(i0)')6+floor(log10(val))
elseif(val.lt.-1)then
write(fstring(3:5),'(i0)')7+floor(log10(-val))
elseif(val.ge.0)
write(fstring(3:5),'(i0)')6
else
write(fstring(3:5),'(i0)')7
endif
write(*,fstring)val
just for fun with modern fortran that supports character functions you can roll that up in a function and end up with a construct like this:
write(*,'('//fstring(val1)//','//fstring(val2)//')')val1,val2
Can any one please help me how to get float value as it is from text box
for Ex: I have entered 40.7
rateField=[[rateField text] floatValue];
I am getting rateField value as 40.7000008 but I want 40.7 only.
please help me.
thanks in advance
Thanks Every body,
I tried all the possibilities but I am not able to get what I want. I am not looking to print the value to convert into string.I want to use that value for computation. If i use Number Formatter again when i am converting from number to float it is giving same problem.So i want float value only but it should be whatever i have given in the text box it should not be padded with any values.This is my requirement.Please help me.
thanks®ards Balu
Thanks Every body,
I tried all the possibilities but I am not able to get what I want. I am not looking to print the value to convert into string.I want to use that value for computation. If i use Number Formatter again when i am converting from number to float it is giving same problem.So i want float value only but it should be whatever i have given in the text box it should not be padded with any values.This is my requirement.Please help me.
thanks®ards
Balu
This is ok. There is not guaranteed that you will get 40.7 if you will use even double.
If you want to output 40.7 you can use %.1f or NSNumberFormatter
Try using a double instead. Usually solves that issue. Has to do with the storage precision.
double dbl = [rateField.text doubleValue];
When using floating point numbers, these things can happen because of the way the numbers are stored in binary format in the computers memory.
It's similar to the way 1/3 = 0.33333333333333... in decimal numbers.
The best way to deal with this is to use number formatters in the textbox that displays the value.
You are already resolved float value.
Floating point numbers have limited precision. Although it depends on
the system, float relative error due to rounding will be around 1.1e-8
Non elementary arithmetic operations may give larger errors, and, of
course, error progragation must be considered when several operations
are compounded.
Additionally, rational numbers that are exactly representable as
floating point numbers in base 10, like 0.1 or 0.7, do not have an
exact representation as floating point numbers in base 2, which is
used internally, no matter the size of the mantissa. Hence, they
cannot be converted into their internal binary counterparts without a
small loss of precision. This can lead to confusing results: for
example, floor((0.1+0.7)*10) will usually return 7 instead of the
expected 8, since the internal representation will be something like
7.9999999999999991118....
So if you're using those numbers for output, you should use some rounding mechanism, even for double values.