BASH: Count identical lines

I have a file that contains:

VoicemailButtonTest
VoicemailButtonTest
VoicemailButtonTest
VoicemailButtonTest
VoicemailButtonTest
VoiceMailConfig60CharsTest
VoicemailDefaultTypeTest
VoiceMailIconSelectableTest
VoiceMailIconSelectableTest
VoiceMailIconSelectableTest
VoiceMailIconSelectableTest
VoiceMailIconSelectableTest
VoicemailSettingsFromMessageModeScreenTest
VoicemailSettingsFromMessageModeScreenTest
VoicemailSettingsTest
VoicemailSettingsTest
VoicemailSettingsTest
VoicemailSettingsTest
VoicemailSettingsTest
VoicemailSettingsTest
VoicemailSettingsTest

How do I replace the duplicate lines with counts:

VoicemailButtonTest (5)
VoiceMailConfig60CharsTest (1)
VoicemailDefaultTypeTest (1)
VoiceMailIconSelectableTest (5)
VoicemailSettingsFromMessageModeScreenTest (2)
VoicemailSettingsTest (7)

I placing the pair into an associative array. I tried using 'read' inside a 'while' statement, but the array gets lost. Here's my attempt:

unset line
tests=$(cat file.log)
echo "$tests" | 
    while read l; do 
        if [ "$l" == "${line}" ]; then
            let cnt++;
        else
            echo "${line} (${cnt})"
            line=${l}
            cnt=1
        fi
        export run_suites
    done

Solution

Assuming the formatting of the output doesn't exactly have to match

VoicemailButtonTest (5)
VoiceMailConfig60CharsTest (1)
VoicemailDefaultTypeTest (1)
VoiceMailIconSelectableTest (5)
VoicemailSettingsFromMessageModeScreenTest (2)
VoicemailSettingsTest (7)

you can just use

sort <input_file> | uniq -c

If you need the output to exactly match what you posted, you can use

awk '{duplicates[$1]++} END{for (ind in duplicates) {print ind,"("duplicates[ind]")"}}' <input_file>

Edit: Posted just after anubhava's answer... but leaving (unless people suggest I delete) because of the addition of the sort command.