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I know a website offering a test for knowing how many words you know in English.

http://testyourvocab.com/

I would like to find something similar for the German language.

I've made some searches but I couldn't find anything. For finding the English one, it was enough to search for "test English words".

Do you know any?

EDIT

I've found this site by making this Google search. But I would like to find one for German..

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    Does it really help you to know a definite number? Test yourself by reading original texts in various levels from fairy tales to novels and newspapers, then you will know whether your vocabulary is sufficient or not.
    – rogermue
    Commented Apr 16, 2014 at 9:40
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    Thanks, I'm studying german with an apps which is focused on vocabulary. I should know between 700 and 1500 words. But my grammar is really bad.. and also the real application of these words.. I cannot read a comic of Mikey Mouse yes.. I read one word and I "decode it" and continue..
    – Revious
    Commented Apr 16, 2014 at 10:01
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    A method I would consider laborious, uninteresting and not very efficient. I never learnt a language this way.
    – rogermue
    Commented Apr 16, 2014 at 10:15
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    Which way do you use? (I use space repeated en.wikipedia.org/wiki/Spaced_repetition)
    – Revious
    Commented Apr 16, 2014 at 10:29
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    Never heard of such theories, and there are hundreds. For beginning I choose a simple one-volume course book with a beginner's dictionary and and beginner's grammar. Then I begin reading simple texts such as children's stories I find in public libraries. Those stories are simple, easy to understand as they are illustrated, and the stories are good.
    – rogermue
    Commented Apr 16, 2014 at 10:39

3 Answers 3

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Here's a German vocab test site very similar to the one in the question: https://wortschatz.tk/ Sounds like what you're looking for.

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+50

I'll recommend the test by the "Institut für Testforschung und Testentwicklung" in Leipzig:

http://www.itt-leipzig.de/static/startseiteeng.html

You can find alternatives if you search the Internet for "Wortschatztest".

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    I have to recommend that this answer receive the bounty. It's a solid answer, and better than anything I found while googling it.
    – emaltman
    Commented Apr 26, 2014 at 1:38
  • @emaltman: it's indeed really interesting and good but I can do it only in english since to guess the "hello" obvious word I have to been able to understand the many not obvious words like: "used to express a greeting, answer a telephone, or attract attention."
    – Revious
    Commented Apr 26, 2014 at 14:40
  • @Revious, you may be right, but only for the "lower end" of the tests. The only drawback I found was that near hits or valid (although less likely) alternatives in the "productive" part of the tests can not be evaluated as such.
    – Walter Tross
    Commented Apr 29, 2014 at 21:39
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    The other problem I found was in the "Italian - receptive" test: I scored 29/30, 29/30, 30/30, 29/30, 27/30, although Italian is my first language, and I have no problem at all with it. I scored much better in German, which is my second language, and even in English.
    – Walter Tross
    Commented Apr 29, 2014 at 22:06
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    Update to my previous comment: I repeated the "Italian - receptive" test with great care, and scored 29/30, 29/30, 30/30, 30/30, 28/30. Several questions made me scratch my head. This particular test is not reliable for sure, which of course causes the rieliability of the others to be questionable too.
    – Walter Tross
    Commented Apr 30, 2014 at 7:48
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Depending on your vocabulary, get an Open dictionary. Free software comes with such dictionaries.

Get the number of entries in the dictionary. 

wc -l /usr/share/hunspell/de_DE.dic
72374 /usr/share/hunspell/de_DE.dic

Now take 100 words by random from the dictionary, and count how much you know. This schould be a good estimate. For higher acccuracy take 1000 samples and divide the result by 10. 

for dummy in {1..100}; do  z=$((RANDOM*RANDOM%72374)); sed -n ${z}p /usr/share/hunspell/de_DE.dic; done

The result (17% = 17/100) multiplied with size of dict (17*72374/100)=12303 to get the absolute result.

Since you can concatenate substantives in German (Holzbein, Eichenholzbein, Eichenholzbeinholzvorrat) the result is open for interpretation.

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  • @User_unknown: you know that it's a really nice idea.. maybe hard to realize (I'm a developer bit also a bit lazy) but it's nice!
    – Revious
    Commented Apr 26, 2014 at 14:04
  • I had a look at /usr/share/hunspell/en_US.dic, and found that ca. 27% of it consists of proper nouns (which you can tell from the rest because of the capital first letter). I wouldn't know how to separate proper nouns from the rest in de_DE.dic, unfortunately. BTW, in .dic files the number of entries is the content of the first line.
    – Walter Tross
    Commented Apr 29, 2014 at 21:28
  • With egrep -n "^a" /usr/share/hunspell/en_US.dic | head -n 1 you'll get the line with the first a in the first column (16709:a in my case). I'm not sure whether that is what you're after.
    – user unknown
    Commented Apr 30, 2014 at 1:45

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