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a correction to my message

This statement
puts the A column value in the corresponding row in column B if that value
is less than 28, otherwise it puts 0 in the column B row.

Should instead read as follows
puts the A column value in the corresponding row in column B if that value
is less than 29, otherwise it puts 0 in the column B row.

or, to state it the way the formula is written
puts 0 in the corresponding row in column B if the value in column A is
greater than 28, otherwise it puts the A row value in the corresponding B
row.

On Tue, Oct 20, 2020 at 7:03 AM Alan B <aboba0@gmail.com> wrote:

Dave,

Not sure I'm following your intention but...

This formula:
{=IF(A1:A13>28,0,A1:A13)}

puts the A column value in the corresponding row in column B if that value
is less than 28, otherwise it puts 0 in the column B row.

Note the curly braces surrounding the formula. These cannot be typed in.
The formula will not work if they are. To enter this formula, type it in
without the curly braces and then use Ctrl+Shift+Enter to complete the
formula.

Ctrl+Shift+Enter are the magic keys to create an array formula. Once
formula entry is completed with those keys you will see the formula has the
curly braces, values in A*n* will be in B*n* if they are less than 28,
otherwise 0 will appear in B*n*.

On Mon, Oct 19, 2020 at 11:07 PM Dave Stevens <geek@uniserve.com> wrote:

I have a column of 365 daily values, non-negative integers, (for a
calendar year)

I want a columns (column 2) just like that but with some cells (start
and end row specified) to be zeroed if they are too large.

so...

1       1
3       3
21      21
29      0
22      22
0       0

All values in column 1 to be copied to column 2 and if
greater than 28 replaced by zero.

The help page gives a highly relevant IF function example that almost
works. I just don't see how to write the value of the column 1 contents
rather than the cell reference. The modified example is
'=IF(A1:A13>28,0,"too small")'


at first I tried changing "too small" to A1 but I get "A1" in
column 2 rather than a value.

I don't see the syntax to get column 1's value rather than its
reference.

Anyone?

Dave


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--
Alan Boba
CISSP, CCENT, ITIL v3 Foundations 2011



-- 
Alan Boba
CISSP, CCENT, ITIL v3 Foundations 2011

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