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03.tex
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03.tex
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\maketitle
\section{Data Structures}\label{data-structures}
In simple terms, It is the the collection or group of data in a
particular structure.
\subsection{Lists}\label{lists}
Lists are the most commonly used data structure. Think of it as a
sequence of data that is enclosed in square brackets and data are
separated by a comma. Each of these data can be accessed by calling it's
index value.
Lists are declared by just equating a variable to `{[} {]}' or list.
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\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}2}]:} \PY{k}{print} \PY{n+nb}{type}\PY{p}{(}\PY{n}{a}\PY{p}{)}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
<type 'list'>
\end{Verbatim}
One can directly assign the sequence of data to a list x as shown.
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}3}]:} \PY{n}{x} \PY{o}{=} \PY{p}{[}\PY{l+s}{\PYZsq{}}\PY{l+s}{apple}\PY{l+s}{\PYZsq{}}\PY{p}{,} \PY{l+s}{\PYZsq{}}\PY{l+s}{orange}\PY{l+s}{\PYZsq{}}\PY{p}{]}
\end{Verbatim}
\subsubsection{Indexing}\label{indexing}
In python, Indexing starts from 0. Thus now the list x, which has two
elements will have apple at 0 index and orange at 1 index.
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}4}]:} \PY{n}{x}\PY{p}{[}\PY{l+m+mi}{0}\PY{p}{]}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{outcolor}Out[{\color{outcolor}4}]:} 'apple'
\end{Verbatim}
Indexing can also be done in reverse order. That is the last element can
be accessed first. Here, indexing starts from -1. Thus index value -1
will be orange and index -2 will be apple.
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}5}]:} \PY{n}{x}\PY{p}{[}\PY{o}{\PYZhy{}}\PY{l+m+mi}{1}\PY{p}{]}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{outcolor}Out[{\color{outcolor}5}]:} 'orange'
\end{Verbatim}
As you might have already guessed, x{[}0{]} = x{[}-2{]}, x{[}1{]} =
x{[}-1{]}. This concept can be extended towards lists with more many
elements.
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}6}]:} \PY{n}{y} \PY{o}{=} \PY{p}{[}\PY{l+s}{\PYZsq{}}\PY{l+s}{carrot}\PY{l+s}{\PYZsq{}}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{potato}\PY{l+s}{\PYZsq{}}\PY{p}{]}
\end{Verbatim}
Here we have declared two lists x and y each containing its own data.
Now, these two lists can again be put into another list say z which will
have it's data as two lists. This list inside a list is called as nested
lists and is how an array would be declared which we will see later.
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}7}]:} \PY{n}{z} \PY{o}{=} \PY{p}{[}\PY{n}{x}\PY{p}{,}\PY{n}{y}\PY{p}{]}
\PY{k}{print} \PY{n}{z}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
[['apple', 'orange'], ['carrot', 'potato']]
\end{Verbatim}
Indexing in nested lists can be quite confusing if you do not understand
how indexing works in python. So let us break it down and then arrive at
a conclusion.
Let us access the data `apple' in the above nested list. First, at index
0 there is a list {[}`apple',`orange'{]} and at index 1 there is another
list {[}`carrot',`potato'{]}. Hence z{[}0{]} should give us the first
list which contains `apple'.
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}8}]:} \PY{n}{z1} \PY{o}{=} \PY{n}{z}\PY{p}{[}\PY{l+m+mi}{0}\PY{p}{]}
\PY{k}{print} \PY{n}{z1}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
['apple', 'orange']
\end{Verbatim}
Now observe that z1 is not at all a nested list thus to access `apple',
z1 should be indexed at 0.
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}9}]:} \PY{n}{z1}\PY{p}{[}\PY{l+m+mi}{0}\PY{p}{]}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{outcolor}Out[{\color{outcolor}9}]:} 'apple'
\end{Verbatim}
Instead of doing the above, In python, you can access `apple' by just
writing the index values each time side by side.
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}10}]:} \PY{n}{z}\PY{p}{[}\PY{l+m+mi}{0}\PY{p}{]}\PY{p}{[}\PY{l+m+mi}{0}\PY{p}{]}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{outcolor}Out[{\color{outcolor}10}]:} 'apple'
\end{Verbatim}
If there was a list inside a list inside a list then you can access the
innermost value by executing z{[} {]}{[} {]}{[} {]}.
\subsubsection{Slicing}\label{slicing}
Indexing was only limited to accessing a single element, Slicing on the
other hand is accessing a sequence of data inside the list. In other
words ``slicing'' the list.
Slicing is done by defining the index values of the first element and
the last element from the parent list that is required in the sliced
list. It is written as parentlist{[} a : b {]} where a,b are the index
values from the parent list. If a or b is not defined then the index
value is considered to be the first value for a if a is not defined and
the last value for b when b is not defined.
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}11}]:} \PY{n}{num} \PY{o}{=} \PY{p}{[}\PY{l+m+mi}{0}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{4}\PY{p}{,}\PY{l+m+mi}{5}\PY{p}{,}\PY{l+m+mi}{6}\PY{p}{,}\PY{l+m+mi}{7}\PY{p}{,}\PY{l+m+mi}{8}\PY{p}{,}\PY{l+m+mi}{9}\PY{p}{]}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}12}]:} \PY{k}{print} \PY{n}{num}\PY{p}{[}\PY{l+m+mi}{0}\PY{p}{:}\PY{l+m+mi}{4}\PY{p}{]}
\PY{k}{print} \PY{n}{num}\PY{p}{[}\PY{l+m+mi}{4}\PY{p}{:}\PY{p}{]}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
[0, 1, 2, 3]
[4, 5, 6, 7, 8, 9]
\end{Verbatim}
You can also slice a parent list with a fixed length or step length.
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}13}]:} \PY{n}{num}\PY{p}{[}\PY{p}{:}\PY{l+m+mi}{9}\PY{p}{:}\PY{l+m+mi}{3}\PY{p}{]}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{outcolor}Out[{\color{outcolor}13}]:} [0, 3, 6]
\end{Verbatim}
\subsubsection{Built in List Functions}\label{built-in-list-functions}
To find the length of the list or the number of elements in a list,
\textbf{len( )} is used.
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}14}]:} \PY{n+nb}{len}\PY{p}{(}\PY{n}{num}\PY{p}{)}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{outcolor}Out[{\color{outcolor}14}]:} 10
\end{Verbatim}
If the list consists of all integer elements then \textbf{min( )} and
\textbf{max( )} gives the minimum and maximum value in the list.
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}15}]:} \PY{n+nb}{min}\PY{p}{(}\PY{n}{num}\PY{p}{)}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{outcolor}Out[{\color{outcolor}15}]:} 0
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}16}]:} \PY{n+nb}{max}\PY{p}{(}\PY{n}{num}\PY{p}{)}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{outcolor}Out[{\color{outcolor}16}]:} 9
\end{Verbatim}
Lists can be concatenated by adding, `+' them. The resultant list will
contain all the elements of the lists that were added. The resultant
list will not be a nested list.
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}17}]:} \PY{p}{[}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{]} \PY{o}{+} \PY{p}{[}\PY{l+m+mi}{5}\PY{p}{,}\PY{l+m+mi}{4}\PY{p}{,}\PY{l+m+mi}{7}\PY{p}{]}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{outcolor}Out[{\color{outcolor}17}]:} [1, 2, 3, 5, 4, 7]
\end{Verbatim}
There might arise a requirement where you might need to check if a
particular element is there in a predefined list. Consider the below
list.
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}18}]:} \PY{n}{names} \PY{o}{=} \PY{p}{[}\PY{l+s}{\PYZsq{}}\PY{l+s}{Earth}\PY{l+s}{\PYZsq{}}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{Air}\PY{l+s}{\PYZsq{}}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{Fire}\PY{l+s}{\PYZsq{}}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{Water}\PY{l+s}{\PYZsq{}}\PY{p}{]}
\end{Verbatim}
To check if `Fire' and `Rajath' is present in the list names. A
conventional approach would be to use a for loop and iterate over the
list and use the if condition. But in python you can use `a in b'
concept which would return `True' if a is present in b and `False' if
not.
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}19}]:} \PY{l+s}{\PYZsq{}}\PY{l+s}{Fire}\PY{l+s}{\PYZsq{}} \PY{o+ow}{in} \PY{n}{names}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{outcolor}Out[{\color{outcolor}19}]:} True
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}20}]:} \PY{l+s}{\PYZsq{}}\PY{l+s}{Rajath}\PY{l+s}{\PYZsq{}} \PY{o+ow}{in} \PY{n}{names}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{outcolor}Out[{\color{outcolor}20}]:} False
\end{Verbatim}
In a list with elements as string, \textbf{max( )} and \textbf{min( )}
is applicable. \textbf{max( )} would return a string element whose ASCII
value is the highest and the lowest when \textbf{min( )} is used. Note
that only the first index of each element is considered each time and if
they value is the same then second index considered so on and so forth.
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}21}]:} \PY{n}{mlist} \PY{o}{=} \PY{p}{[}\PY{l+s}{\PYZsq{}}\PY{l+s}{bzaa}\PY{l+s}{\PYZsq{}}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{ds}\PY{l+s}{\PYZsq{}}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{nc}\PY{l+s}{\PYZsq{}}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{az}\PY{l+s}{\PYZsq{}}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{z}\PY{l+s}{\PYZsq{}}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{klm}\PY{l+s}{\PYZsq{}}\PY{p}{]}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}22}]:} \PY{k}{print} \PY{n+nb}{max}\PY{p}{(}\PY{n}{mlist}\PY{p}{)}
\PY{k}{print} \PY{n+nb}{min}\PY{p}{(}\PY{n}{mlist}\PY{p}{)}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
z
az
\end{Verbatim}
Here the first index of each element is considered and thus z has the
highest ASCII value thus it is returned and minimum ASCII is a. But what
if numbers are declared as strings?
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}23}]:} \PY{n}{nlist} \PY{o}{=} \PY{p}{[}\PY{l+s}{\PYZsq{}}\PY{l+s}{1}\PY{l+s}{\PYZsq{}}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{94}\PY{l+s}{\PYZsq{}}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{93}\PY{l+s}{\PYZsq{}}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{1000}\PY{l+s}{\PYZsq{}}\PY{p}{]}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}24}]:} \PY{k}{print} \PY{n+nb}{max}\PY{p}{(}\PY{n}{nlist}\PY{p}{)}
\PY{k}{print} \PY{n+nb}{min}\PY{p}{(}\PY{n}{nlist}\PY{p}{)}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
94
1
\end{Verbatim}
Even if the numbers are declared in a string the first index of each
element is considered and the maximum and minimum values are returned
accordingly.
But if you want to find the \textbf{max( )} string element based on the
length of the string then another parameter `key=len' is declared inside
the \textbf{max( )} and \textbf{min( )} function.
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}25}]:} \PY{k}{print} \PY{n+nb}{max}\PY{p}{(}\PY{n}{names}\PY{p}{,} \PY{n}{key}\PY{o}{=}\PY{n+nb}{len}\PY{p}{)}
\PY{k}{print} \PY{n+nb}{min}\PY{p}{(}\PY{n}{names}\PY{p}{,} \PY{n}{key}\PY{o}{=}\PY{n+nb}{len}\PY{p}{)}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
Earth
Air
\end{Verbatim}
But even `Water' has length 5. \textbf{max()} or \textbf{min()} function
returns the first element when there are two or more elements with the
same length.
Any other built in function can be used or lambda function (will be
discussed later) in place of len.
A string can be converted into a list by using the \textbf{list()}
function.
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}26}]:} \PY{n+nb}{list}\PY{p}{(}\PY{l+s}{\PYZsq{}}\PY{l+s}{hello}\PY{l+s}{\PYZsq{}}\PY{p}{)}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{outcolor}Out[{\color{outcolor}26}]:} ['h', 'e', 'l', 'l', 'o']
\end{Verbatim}
\textbf{append( )} is used to add a element at the end of the list.
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}27}]:} \PY{n}{lst} \PY{o}{=} \PY{p}{[}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{4}\PY{p}{,}\PY{l+m+mi}{8}\PY{p}{,}\PY{l+m+mi}{7}\PY{p}{]}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}28}]:} \PY{n}{lst}\PY{o}{.}\PY{n}{append}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}
\PY{k}{print} \PY{n}{lst}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
[1, 1, 4, 8, 7, 1]
\end{Verbatim}
\textbf{count( )} is used to count the number of a particular element
that is present in the list.
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}29}]:} \PY{n}{lst}\PY{o}{.}\PY{n}{count}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{outcolor}Out[{\color{outcolor}29}]:} 3
\end{Verbatim}
\textbf{append( )} function can also be used to add a entire list at the
end. Observe that the resultant list becomes a nested list.
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}30}]:} \PY{n}{lst1} \PY{o}{=} \PY{p}{[}\PY{l+m+mi}{5}\PY{p}{,}\PY{l+m+mi}{4}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{8}\PY{p}{]}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}31}]:} \PY{n}{lst}\PY{o}{.}\PY{n}{append}\PY{p}{(}\PY{n}{lst1}\PY{p}{)}
\PY{k}{print} \PY{n}{lst}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
[1, 1, 4, 8, 7, 1, [5, 4, 2, 8]]
\end{Verbatim}
But if nested list is not what is desired then \textbf{extend( )}
function can be used.
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}32}]:} \PY{n}{lst}\PY{o}{.}\PY{n}{extend}\PY{p}{(}\PY{n}{lst1}\PY{p}{)}
\PY{k}{print} \PY{n}{lst}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
[1, 1, 4, 8, 7, 1, [5, 4, 2, 8], 5, 4, 2, 8]
\end{Verbatim}
\textbf{index( )} is used to find the index value of a particular
element. Note that if there are multiple elements of the same value then
the first index value of that element is returned.
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}33}]:} \PY{n}{lst}\PY{o}{.}\PY{n}{index}\PY{p}{(}\PY{l+m+mi}{1}\PY{p}{)}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{outcolor}Out[{\color{outcolor}33}]:} 0
\end{Verbatim}
\textbf{insert(x,y)} is used to insert a element y at a specified index
value x. \textbf{append( )} function made it only possible to insert at
the end.
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}34}]:} \PY{n}{lst}\PY{o}{.}\PY{n}{insert}\PY{p}{(}\PY{l+m+mi}{5}\PY{p}{,} \PY{l+s}{\PYZsq{}}\PY{l+s}{name}\PY{l+s}{\PYZsq{}}\PY{p}{)}
\PY{k}{print} \PY{n}{lst}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
[1, 1, 4, 8, 7, 'name', 1, [5, 4, 2, 8], 5, 4, 2, 8]
\end{Verbatim}
\textbf{insert(x,y)} inserts but does not replace element. If you want
to replace the element with another element you simply assign the value
to that particular index.
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}35}]:} \PY{n}{lst}\PY{p}{[}\PY{l+m+mi}{5}\PY{p}{]} \PY{o}{=} \PY{l+s}{\PYZsq{}}\PY{l+s}{Python}\PY{l+s}{\PYZsq{}}
\PY{k}{print} \PY{n}{lst}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
[1, 1, 4, 8, 7, 'Python', 1, [5, 4, 2, 8], 5, 4, 2, 8]
\end{Verbatim}
\textbf{pop( )} function return the last element in the list. This is
similar to the operation of a stack. Hence it wouldn't be wrong to tell
that lists can be used as a stack.
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}36}]:} \PY{n}{lst}\PY{o}{.}\PY{n}{pop}\PY{p}{(}\PY{p}{)}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{outcolor}Out[{\color{outcolor}36}]:} 8
\end{Verbatim}
Index value can be specified to pop a ceratin element corresponding to
that index value.
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}37}]:} \PY{n}{lst}\PY{o}{.}\PY{n}{pop}\PY{p}{(}\PY{l+m+mi}{0}\PY{p}{)}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{outcolor}Out[{\color{outcolor}37}]:} 1
\end{Verbatim}
\textbf{pop( )} is used to remove element based on it's index value
which can be assigned to a variable. One can also remove element by
specifying the element itself using the \textbf{remove( )} function.
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}38}]:} \PY{n}{lst}\PY{o}{.}\PY{n}{remove}\PY{p}{(}\PY{l+s}{\PYZsq{}}\PY{l+s}{Python}\PY{l+s}{\PYZsq{}}\PY{p}{)}
\PY{k}{print} \PY{n}{lst}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
[1, 4, 8, 7, 1, [5, 4, 2, 8], 5, 4, 2]
\end{Verbatim}
Alternative to \textbf{remove} function but with using index value is
\textbf{del}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}39}]:} \PY{k}{del} \PY{n}{lst}\PY{p}{[}\PY{l+m+mi}{1}\PY{p}{]}
\PY{k}{print} \PY{n}{lst}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
[1, 8, 7, 1, [5, 4, 2, 8], 5, 4, 2]
\end{Verbatim}
The entire elements present in the list can be reversed by using the
\textbf{reverse()} function.
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}40}]:} \PY{n}{lst}\PY{o}{.}\PY{n}{reverse}\PY{p}{(}\PY{p}{)}
\PY{k}{print} \PY{n}{lst}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
[2, 4, 5, [5, 4, 2, 8], 1, 7, 8, 1]
\end{Verbatim}
Note that the nested list {[}5,4,2,8{]} is treated as a single element
of the parent list lst. Thus the elements inside the nested list is not
reversed.
Python offers built in operation \textbf{sort( )} to arrange the
elements in ascending order.
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}41}]:} \PY{n}{lst}\PY{o}{.}\PY{n}{sort}\PY{p}{(}\PY{p}{)}
\PY{k}{print} \PY{n}{lst}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
[1, 1, 2, 4, 5, 7, 8, [5, 4, 2, 8]]
\end{Verbatim}
For descending order, By default the reverse condition will be False for
reverse. Hence changing it to True would arrange the elements in
descending order.
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}42}]:} \PY{n}{lst}\PY{o}{.}\PY{n}{sort}\PY{p}{(}\PY{n}{reverse}\PY{o}{=}\PY{n+nb+bp}{True}\PY{p}{)}
\PY{k}{print} \PY{n}{lst}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
[[5, 4, 2, 8], 8, 7, 5, 4, 2, 1, 1]
\end{Verbatim}
Similarly for lists containing string elements, \textbf{sort( )} would
sort the elements based on it's ASCII value in ascending and by
specifying reverse=True in descending.
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}43}]:} \PY{n}{names}\PY{o}{.}\PY{n}{sort}\PY{p}{(}\PY{p}{)}
\PY{k}{print} \PY{n}{names}
\PY{n}{names}\PY{o}{.}\PY{n}{sort}\PY{p}{(}\PY{n}{reverse}\PY{o}{=}\PY{n+nb+bp}{True}\PY{p}{)}
\PY{k}{print} \PY{n}{names}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
['Air', 'Earth', 'Fire', 'Water']
['Water', 'Fire', 'Earth', 'Air']
\end{Verbatim}
To sort based on length key=len should be specified as shown.
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}44}]:} \PY{n}{names}\PY{o}{.}\PY{n}{sort}\PY{p}{(}\PY{n}{key}\PY{o}{=}\PY{n+nb}{len}\PY{p}{)}
\PY{k}{print} \PY{n}{names}
\PY{n}{names}\PY{o}{.}\PY{n}{sort}\PY{p}{(}\PY{n}{key}\PY{o}{=}\PY{n+nb}{len}\PY{p}{,}\PY{n}{reverse}\PY{o}{=}\PY{n+nb+bp}{True}\PY{p}{)}
\PY{k}{print} \PY{n}{names}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
['Air', 'Fire', 'Water', 'Earth']
['Water', 'Earth', 'Fire', 'Air']
\end{Verbatim}
\subsubsection{Copying a list}\label{copying-a-list}
Most of the new python programmers commit this mistake. Consider the
following,
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}45}]:} \PY{n}{lista}\PY{o}{=} \PY{p}{[}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{4}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{]}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}46}]:} \PY{n}{listb} \PY{o}{=} \PY{n}{lista}
\PY{k}{print} \PY{n}{listb}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
[2, 1, 4, 3]
\end{Verbatim}
Here, We have declared a list, lista = {[}2,1,4,3{]}. This list is
copied to listb by assigning it's value and it get's copied as seen. Now
we perform some random operations on lista.
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}47}]:} \PY{n}{lista}\PY{o}{.}\PY{n}{pop}\PY{p}{(}\PY{p}{)}
\PY{k}{print} \PY{n}{lista}
\PY{n}{lista}\PY{o}{.}\PY{n}{append}\PY{p}{(}\PY{l+m+mi}{9}\PY{p}{)}
\PY{k}{print} \PY{n}{lista}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
[2, 1, 4]
[2, 1, 4, 9]
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}48}]:} \PY{k}{print} \PY{n}{listb}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
[2, 1, 4, 9]
\end{Verbatim}
listb has also changed though no operation has been performed on it.
This is because you have assigned the same memory space of lista to
listb. So how do fix this?
If you recall, in slicing we had seen that parentlist{[}a:b{]} returns a
list from parent list with start index a and end index b and if a and b
is not mentioned then by default it considers the first and last
element. We use the same concept here. By doing so, we are assigning the
data of lista to listb as a variable.
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}49}]:} \PY{n}{lista} \PY{o}{=} \PY{p}{[}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{4}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{]}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}50}]:} \PY{n}{listb} \PY{o}{=} \PY{n}{lista}\PY{p}{[}\PY{p}{:}\PY{p}{]}
\PY{k}{print} \PY{n}{listb}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
[2, 1, 4, 3]
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}51}]:} \PY{n}{lista}\PY{o}{.}\PY{n}{pop}\PY{p}{(}\PY{p}{)}
\PY{k}{print} \PY{n}{lista}
\PY{n}{lista}\PY{o}{.}\PY{n}{append}\PY{p}{(}\PY{l+m+mi}{9}\PY{p}{)}
\PY{k}{print} \PY{n}{lista}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
[2, 1, 4]
[2, 1, 4, 9]
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}52}]:} \PY{k}{print} \PY{n}{listb}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
[2, 1, 4, 3]
\end{Verbatim}
\subsection{Tuples}\label{tuples}
Tuples are similar to lists but only big difference is the elements
inside a list can be changed but in tuple it cannot be changed. Think of
tuples as something which has to be True for a particular something and
cannot be True for no other values. For better understanding, Recall
\textbf{divmod()} function.
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}53}]:} \PY{n}{xyz} \PY{o}{=} \PY{n+nb}{divmod}\PY{p}{(}\PY{l+m+mi}{10}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{)}
\PY{k}{print} \PY{n}{xyz}
\PY{k}{print} \PY{n+nb}{type}\PY{p}{(}\PY{n}{xyz}\PY{p}{)}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
(3, 1)
<type 'tuple'>
\end{Verbatim}
Here the quotient has to be 3 and the remainder has to be 1. These
values cannot be changed whatsoever when 10 is divided by 3. Hence
divmod returns these values in a tuple.
To define a tuple, A variable is assigned to paranthesis ( ) or tuple(
).
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}54}]:} \PY{n}{tup} \PY{o}{=} \PY{p}{(}\PY{p}{)}
\PY{n}{tup2} \PY{o}{=} \PY{n+nb}{tuple}\PY{p}{(}\PY{p}{)}
\end{Verbatim}
If you want to directly declare a tuple it can be done by using a comma
at the end of the data.
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}55}]:} \PY{l+m+mi}{27}\PY{p}{,}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{outcolor}Out[{\color{outcolor}55}]:} (27,)
\end{Verbatim}
27 when multiplied by 2 yields 54, But when multiplied with a tuple the
data is repeated twice.
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}56}]:} \PY{l+m+mi}{2}\PY{o}{*}\PY{p}{(}\PY{l+m+mi}{27}\PY{p}{,}\PY{p}{)}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{outcolor}Out[{\color{outcolor}56}]:} (27, 27)
\end{Verbatim}
Values can be assigned while declaring a tuple. It takes a list as input
and converts it into a tuple or it takes a string and converts it into a
tuple.
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}57}]:} \PY{n}{tup3} \PY{o}{=} \PY{n+nb}{tuple}\PY{p}{(}\PY{p}{[}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{]}\PY{p}{)}
\PY{k}{print} \PY{n}{tup3}
\PY{n}{tup4} \PY{o}{=} \PY{n+nb}{tuple}\PY{p}{(}\PY{l+s}{\PYZsq{}}\PY{l+s}{Hello}\PY{l+s}{\PYZsq{}}\PY{p}{)}
\PY{k}{print} \PY{n}{tup4}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
(1, 2, 3)
('H', 'e', 'l', 'l', 'o')
\end{Verbatim}
It follows the same indexing and slicing as Lists.
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}58}]:} \PY{k}{print} \PY{n}{tup3}\PY{p}{[}\PY{l+m+mi}{1}\PY{p}{]}
\PY{n}{tup5} \PY{o}{=} \PY{n}{tup4}\PY{p}{[}\PY{p}{:}\PY{l+m+mi}{3}\PY{p}{]}
\PY{k}{print} \PY{n}{tup5}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
2
('H', 'e', 'l')
\end{Verbatim}
\subsubsection{Mapping one tuple to
another}\label{mapping-one-tuple-to-another}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}59}]:} \PY{p}{(}\PY{n}{a}\PY{p}{,}\PY{n}{b}\PY{p}{,}\PY{n}{c}\PY{p}{)}\PY{o}{=} \PY{p}{(}\PY{l+s}{\PYZsq{}}\PY{l+s}{alpha}\PY{l+s}{\PYZsq{}}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{beta}\PY{l+s}{\PYZsq{}}\PY{p}{,}\PY{l+s}{\PYZsq{}}\PY{l+s}{gamma}\PY{l+s}{\PYZsq{}}\PY{p}{)}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}60}]:} \PY{k}{print} \PY{n}{a}\PY{p}{,}\PY{n}{b}\PY{p}{,}\PY{n}{c}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
alpha beta gamma
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}61}]:} \PY{n}{d} \PY{o}{=} \PY{n+nb}{tuple}\PY{p}{(}\PY{l+s}{\PYZsq{}}\PY{l+s}{RajathKumarMP}\PY{l+s}{\PYZsq{}}\PY{p}{)}
\PY{k}{print} \PY{n}{d}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
('R', 'a', 'j', 'a', 't', 'h', 'K', 'u', 'm', 'a', 'r', 'M', 'P')
\end{Verbatim}
\subsubsection{Built In Tuple functions}\label{built-in-tuple-functions}
\textbf{count()} function counts the number of specified element that is
present in the tuple.
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}62}]:} \PY{n}{d}\PY{o}{.}\PY{n}{count}\PY{p}{(}\PY{l+s}{\PYZsq{}}\PY{l+s}{a}\PY{l+s}{\PYZsq{}}\PY{p}{)}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{outcolor}Out[{\color{outcolor}62}]:} 3
\end{Verbatim}
\textbf{index()} function returns the index of the specified element. If
the elements are more than one then the index of the first element of
that specified element is returned
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}63}]:} \PY{n}{d}\PY{o}{.}\PY{n}{index}\PY{p}{(}\PY{l+s}{\PYZsq{}}\PY{l+s}{a}\PY{l+s}{\PYZsq{}}\PY{p}{)}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{outcolor}Out[{\color{outcolor}63}]:} 1
\end{Verbatim}
\subsection{Sets}\label{sets}
Sets are mainly used to eliminate repeated numbers in a sequence/list.
It is also used to perform some standard set operations.
Sets are declared as set() which will initialize a empty set. Also
set({[}sequence{]}) can be executed to declare a set with elements
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}64}]:} \PY{n}{set1} \PY{o}{=} \PY{n+nb}{set}\PY{p}{(}\PY{p}{)}
\PY{k}{print} \PY{n+nb}{type}\PY{p}{(}\PY{n}{set1}\PY{p}{)}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
<type 'set'>
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}65}]:} \PY{n}{set0} \PY{o}{=} \PY{n+nb}{set}\PY{p}{(}\PY{p}{[}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{4}\PY{p}{]}\PY{p}{)}
\PY{k}{print} \PY{n}{set0}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
set([1, 2, 3, 4])
\end{Verbatim}
elements 2,3 which are repeated twice are seen only once. Thus in a set
each element is distinct.
\subsubsection{Built-in Functions}\label{built-in-functions}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}66}]:} \PY{n}{set1} \PY{o}{=} \PY{n+nb}{set}\PY{p}{(}\PY{p}{[}\PY{l+m+mi}{1}\PY{p}{,}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{]}\PY{p}{)}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}67}]:} \PY{n}{set2} \PY{o}{=} \PY{n+nb}{set}\PY{p}{(}\PY{p}{[}\PY{l+m+mi}{2}\PY{p}{,}\PY{l+m+mi}{3}\PY{p}{,}\PY{l+m+mi}{4}\PY{p}{,}\PY{l+m+mi}{5}\PY{p}{]}\PY{p}{)}
\end{Verbatim}
\textbf{union( )} function returns a set which contains all the elements
of both the sets without repition.
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}68}]:} \PY{n}{set1}\PY{o}{.}\PY{n}{union}\PY{p}{(}\PY{n}{set2}\PY{p}{)}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{outcolor}Out[{\color{outcolor}68}]:} \{1, 2, 3, 4, 5\}
\end{Verbatim}
\textbf{add( )} will add a particular element into the set. Note that
the index of the newly added element is arbitrary and can be placed
anywhere not neccessarily in the end.
\begin{Verbatim}[commandchars=\\\{\}]
{\color{incolor}In [{\color{incolor}69}]:} \PY{n}{set1}\PY{o}{.}\PY{n}{add}\PY{p}{(}\PY{l+m+mi}{0}\PY{p}{)}
\PY{n}{set1}
\end{Verbatim}
\begin{Verbatim}[commandchars=\\\{\}]
{\color{outcolor}Out[{\color{outcolor}69}]:} \{0, 1, 2, 3\}
\end{Verbatim}
\textbf{intersection( )} function outputs a set which contains all the