Google's Contacts app is very slow to load when I want to quickly get a friend's phone number.

Is there a quicker/lighter contacts app?
I have found nothing fast enough in the Market.

My dream app:

  • Loads the list of ~2000 contacts in less than ~3 seconds.
  • No need for picture/details in the main list, just name.
  • Can import contacts list from vcard or similar.
link|improve this question

64% accept rate
Have you looked in the Market? Why isn't anything there viable? – Al Everett Aug 4 '11 at 12:57
If you're just talking about one or two people, you could always just create a shortcut to their contact information on your home screen. – Al Everett Aug 4 '11 at 12:58
@Al: Thanks for the shortcut tip! Unfortunately, I am contacting many different people, so shortcuts won't do in this case. – Nicolas Raoul Aug 5 '11 at 1:50
@Al: Yes I have looked in the Market, all of the apps I have tried are even slower than Google's, because they all show details, picture, etc in the main list, making it very slow to load. – Nicolas Raoul Aug 5 '11 at 1:55
1  
Could someone please answer this question? I too am looking for this. I kinda liked Sony Ericsson in this respect. You could locate any contact immediately. – Mugen Oct 16 '11 at 18:12
show 2 more comments
feedback

1 Answer

up vote 1 down vote accepted

I will recommend Dialer One. I am 99% sure it can show contacts without pictures. It has lots of customization options, t9 search in contacts in slew of different languages. I have 900+ contacts and is really fast, I hope it works that way for 2000 contacts. Be sure to fiddle with the options after giving it a try.

link|improve this answer
Dialer One is wonderful indeed. To make it even better, it seems to be Open Source. Developer website is code.google.com/p/dialerone but unfortunately that project gives error 403. Anyone knows where to find the source code? – Nicolas Raoul Feb 15 at 2:36
feedback

Your Answer

 
or
required, but never shown

Not the answer you're looking for? Browse other questions tagged or ask your own question.