Replies

But did you see Rory’s one, no selecting, no looping!

You were just soooo close!

it needs to be this:

With Selection.Tables(1)
   .AutoFitBehavior wdAutoFitContent
   .AllowAutoFit = True
End With

otherwise all it does is work on the first table in the document instead of the current one.

Well sorta. My post did actually say “short of selecting the whole table” as normally in VBA you can achieve most things without selecting.

This is quite similar to what I had, except I also had a bit to restore the original selection, and and you missed the preferred width setting for the table itself thus:

Sub AutoFitTable()
    Set orig = Selection.Range
    With Selection.Tables(1)
        .AllowAutoFit = False
        .PreferredWidthType = wdPreferredWidthAuto
        .Select
        Selection.Cells.PreferredWidthType = wdPreferredWidthAuto
        Selection.Columns.PreferredWidthType = wdPreferredWidthAuto
    End With
    orig.Select
End Sub

I’ve never really played around with table styles. I’ll take a look, thanks.

However, very slow is relative. The macro does 10 tables in about a second which is quick enough for me.

Yeah I got that,
but this is the same table. All the other properties seemed to get passed OK, but that one did not.

Thanks Hans,
I still wonder why it was only that one property that was missing when it was failing.

Wow, it’s as easy as that!

I’ve been on this for an hour.

Would would all of us loungers do without HansV?

Thanks you sdo much once again Hans.

I had already tried that (well it was actually a 5 row table that I pasted below). The table and cell widths seemed to be correct, but the formatting was still as it had come from Excel (not too much of a problem), but the real problem with that method was that the table columns were offset from the original ones and I can’t sem to line ’em up again.

Attached is a picture of what I ended up with.

Thanks for that.

I can’t seem to see any difference when I change the scale factors. I tried changing them from 800, 600 to 8, 6 and the output file still appeard to be exactly the same.

What difference should I expect to see?

Thanks for all the replies guys.

In this application the cells that I am interested in have a formula that return text or “”. Sad @#$!% that I am, I figured out whilst trying to get to sleep last night that =COUNTA(A:A,”?*”) should get me what I need for this. However I shall store Jezza’s and Aladin’s ideas away for future use.

Thanks once again.

[indent]

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=sum(if(A$1:A$35 “”,1)

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[/indent] That is equivalent to =SUMPRODUCT((A1:A35″”)*1) since SUMPRODUCT is an implicit array function.

[indent]

---

=counta(A$1:A$35)

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[/indent] won’t work because the cells are equal to “” (from an if statement) rather than blank.

Wow, thanks for that Hans.

[indent]

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those in Chart3 can only be used to predict the height of a person from his or her shoe size, NOT to predict someone’s shoe size from his/her height

---

[/indent]

But that is where I have the problem. If I predict that for a particular shoe size, someone will be of a particular height, I would feel that I have defined a relationship between the 2 values.

Suppose I predict that someone with a shoe size of 7.35 should be of height 164. So I find a person with a shoe size of 7.35, find that their height is 164 and then say, “because your height is 164, I predict that your shoe size is 6.15”.

It just doesn’t work????

The thing that gets me here is that, whilst they all agree at the mean values, if I look at Chart3 it tells me that for a female of height 164 I should expect to see a shoe size of 7.35, however, looking at the same data on Chart1 it would predict a shoe size of about 6.15.

Ahhhh, thanks for that Hans.

I had asumed that the trend line was the “line of best fit” for the points on the scatter graph.

Is it possible to display such a beast, which would then be independent of the orientation of the graph axes?

TIA
Regards
Paul