BASH：计算相同的行

Question

我有一个文件，其中包含：

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

如何用计数替换重复行：

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

我将这对放入关联数组中。我尝试在“while”语句中使用“read”，但数组丢失了。这是我的尝试：

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

Answer 1

假设输出的格式不必完全匹配

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

你可以使用

sort <input_file> | uniq -c

如果您需要输出与您发布的内容完全匹配，您可以使用

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

编辑：在anubhava的回答之后发布...但由于添加了排序命令而离开（除非人们建议我删除）。