Finding Missing Spaces before Numbers

@Tomek:
Thanks for the insight. Of course: If formatting is not of importance one can always copy the text into, say, notepad++, and perform the search (and replace) operation there. Copy back the text into word afterwards.

@Falk:
For the example I gave in my previous comment, you can search for "alumin*um" but that is dangerous: it will find any text that will start with "alumin" and end with "um" even if the latter is 500 pages down in the document, as long as it did not find aluminum or aluminium earlier. So if your text starts with Silicoaluminate and ends with a word maximum, almost all text may be selected. Grrrr.

You cannot use "alumin?um" to find aluminum, the ? looks for exactly one letter.

@Falk:
"+" doesn't work in Word, you have to specify the number of occurrences in braces.
Also, Word doesn't accept 0 occurrences so for instance you cannot use (alumin)([i]{0,1})(um) to look for aluminum (American spelling) and aluminium (IUPAC spelling) in one search. Pity!

@Tomek:
Ok - in Regular Expressions the "braces" allow you to give the exact number or a range of numbers the searched-for expression shall be present. {0,5} allows from none to five occurrences, {1,} looks for at least one occurrence to infinite number of occurrences. "+" does the exact same thing, but maybe Microsoft didn't import the entire syntax for RegExp...

@Falk:
instead of + (which doesn't work on my machine) you should use {1,}
i.e.,
([A-Z,a-z]{1,})([0-9]{1,})

@Pocok:
Use the above pattern, then \1 as replace pattern. This will remove all digits that follow a string of alphabetic characters.
You could also use:
([A-Z,a-z])([0-9]{1,}) combine with \1. This will remove all digits after any alphabetic character (may be just the last one of the preceding word).

There are two things you may need to pay attention to:
1. If the digits are followed by a word without intervening space, the two words will be joined without being separated by space. If this is a problem, use "\1 " as the replacement pattern, then separately look for double spaces and replace them with a single one.
2. If you use other languages, you may need to include language-specific characters like á ß ç ñ ü ą ł č, etc in the list of characters to search for.

Pocok - you simply remove the "\2" from the Replace-With text ("\2" in this case standing in for whatever text matched the second set of parentheses in the Find-What text.

Hello, I would like to build upon this: Using the ([A-Z,a-z])([0-9]) I've found what I needed, however I want to REMOVE the digits from this combo, how can I do that in the Replace segment?

If, as the character pattern suggest, we are dealing with 'regular expressions', a '+' after the right brackets ']' respectively might help in selecting the entire number as well as the entire following word in question. Like so:
([A-Z,a-z]+)([0-9]+)
Haven't tried this out though.

Seems that this only works on the letter and number that are next to each other, not on the entire letters string and the entire numbers string. This means that using "\2 \1" instead of "\1 \2" only swaps and separates those middle two characters.

Thanks for hint on the "wildcard" approach - I didn't know about this feature. One little drawback of the presented solution is, that it will not work as desired on all languages. The "regular expression" used in the "Find" textbox only catches normal english alphabet letters from a to z, capitalized or not. In many languages, however, there are letter accents, like the German "Umlaute" (ä,ö,ü) or the French accents (e.g. é, è, ê). Of course, the regular expression can then be expanded to fit the needs of the language used.