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I installed many packages from many PPAs on my system. I want to list all the installed packages which are installed from launchpad PPAs, not repositories.

Is this possible through command-line?

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Possible Duplicate?… – Mitch Apr 13 '14 at 8:42
No.its a different one. – Avinash Raj Apr 13 '14 at 8:46
Not a command line utility, but very useful is the Y PPA Manager. Lists, installs, and removes packages from PPAs. – Jos Apr 13 '14 at 8:52

3 Answers 3

The following command returns the package name and its ppa (if installed from a ppa):

apt-cache policy $(dpkg --get-selections | grep -v deinstall$ | awk '{ print $1 }') | perl -e '@a = <>; $a=join("", @a); $a =~ s/\n(\S)/\n\n$1/g;  @packages = split("\n\n", $a); foreach $p (@packages) {print "$1: $2\n" if $p =~ /^(.*?):.*?500 http:\/\/ppa\.launchpad\.net\/(.*?)\s/s}'


  • dpkg --get-selections gives only the installed packages after grep -v deinstall$
  • awk '{ print $1 }' returns only the package name
  • perl -e '@a = <>; $a=join("", @a)' concatenates all the lines returned by apt-cache policy
  • $a =~ s/\n(\S)/\n\n$1/g; adds a newline between each package section
  • @packages = split("\n\n", $a); is a perl array containing all the packages infos, one package per item.
  • foreach $p (@packages) {print "$1: $2\n" if $p =~ /^(.*?):.*?500 http:\/\/ppa\.launchpad\.net\/(.*?)\s/s} is a loop where the package and the ppa are printed if a ppa with prio 500 is found in the policy.
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Short answer/code, looong time for execution. – Radu Rădeanu Apr 13 '14 at 15:00
@Avinash: did you test my answer? if so could you please consider accept it? Thanks. – Sylvain Pineau Apr 18 '14 at 15:40

In accordance with this answer and this post, you can get a list of all packages from all the PPAs installed on your system using the following bash code:

for APT in $(find /etc/apt/ -name \*.list); do
  grep -o "^deb[a-z0-9\-]\+/[a-z0-9\-]\+" $APT | while read ENTRY ; do
    USER=$(echo $ENTRY | cut -d/ -f4)
    PPA=$(echo $ENTRY | cut -d/ -f5)
    awk '$1 == "Package:" { if (a[$2]++ == 0) print $2; }' /var/lib/apt/lists/*$USER*$PPA*Packages

And in accordance with this answer, you can get a list of all installed packages in your system using:

dpkg --get-selections | grep -v deinstall | cut -f1

Now, let's join these two ideas to get a list of all the packages which are installed from PPAs:

(for APT in $(find /etc/apt/ -name \*.list); do
  grep -o "^deb[a-z0-9\-]\+/[a-z0-9\-]\+" $APT | while read ENTRY ; do
    USER=$(echo $ENTRY | cut -d/ -f4)
    PPA=$(echo $ENTRY | cut -d/ -f5)
    awk '$1 == "Package:" { if (a[$2]++ == 0) print $2; }' /var/lib/apt/lists/*$USER*$PPA*Packages
done; dpkg --get-selections | grep -v deinstall | cut -f1) | sort | awk 'dup[$0]++ == 1'
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The source of an installed package can be checked using apt-cache, for example

$ apt-cache policy oracle-java7-installer

  Installed: 7u51-0~webupd8~7
  Candidate: 7u51-0~webupd8~7
  Version table:
 *** 7u51-0~webupd8~7 0
        500 precise/main i386 Packages
        100 /var/lib/dpkg/status

The output of apt-cache policy <package_name> contains the source.

One can use the following script to obtain the list of packages installed from PPAs.

echo "List of packages which are not installed from Ubuntu repository"
for i in `dpkg -l | grep "^ii" | awk '{print $2}'`
    j=`apt-cache policy "$i" | grep ""` 
    if [ $? -eq 0 ]; then
        echo "$i"
        #echo "$i $j"
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It doesn’t work if you selected a different mirror. For instance I have gir1.2-syncmenu-0.1 500 saucy/main amd64 Packages – Sylvain Pineau Apr 13 '14 at 12:20
In this case google-chrome-stable is not installed from a PPA; it has just a separate repository. – Radu Rădeanu Apr 13 '14 at 13:53
Ok, I saw that. But you came with a really bad example which can make novice users to think that if a package is not from Ubuntu repositories, then the package is from a PPA. The OP's question is about PPAs. – Radu Rădeanu Apr 13 '14 at 14:27
@RaduRădeanu I got your points and Edited my post. you are absolutely correct. – souravc Apr 13 '14 at 14:49
Better now, even if there is a problem with the time for execution which is realy looong. – Radu Rădeanu Apr 13 '14 at 15:05

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