$1.$ To load the data using read$\_$csv for the data files attached $($tips$.$txt or tips.csv$),$ save the files and

path of your own, to include the directory path of the loaded files, like below example of the

saved path:

In: tips$=$ pd$.$read$\_$csv$(\text{'}$c:$/$Users$/$mowaf$/$Data$/$tips$.$txt$\text{'},$ sep$=\text{'},\text{'})$

In: tips$=$ pd$.$read$\_$csv$(\text{'}$c:$/$Users$/$mowaf$/$Data$/$tips$.$csv$\text{'})$

$2.$ To view the file dataset $($first $5$ rows$)$:

In: tips.head$()$

obs totbill tip sex smoker day time size

$0116\mathrm{.}991\mathrm{.}01$ F No Sun Night $2$

$1210\mathrm{.}341\mathrm{.}66$ M No Sun Night $3$

$2321\mathrm{.}013\mathrm{.}50$ M No Sun Night $3$

$3423\mathrm{.}683\mathrm{.}31$ M No Sun Night $2$

$4524\mathrm{.}593\mathrm{.}61$ F No Sun Night $4$

$3.$ To make a stacked bar plot showing the percentage of data points for each party size on each

day, make a cross$-$tabulation by day and party size and sh:

In$[]$: party$\_$count$=$pd$.$crosstab$($tips$[\text{'}$day$\text{'}],$tips$[\text{'}$size$\text{'}])$

In $[]$: party$\_$count

Out:

size $123456$

day

Fri $1161100$

Sat $253181310$

Sun $039151831$

Thu $1484513$

$4.$ To normalize the data, so that each row sums to $1$ and make the plot:

In: party$\_$counts $=$ party$\_$count.loc$[$:$,2$:$5]$

In: party$\_$counts

size $2345$

day

Fri $16110$

Sat $5318131$

Sun $3915183$

Thu $48451$

$5.$ Normalize so that each row sums to $1$ to make a bar plot of the party size $($counts$)$ over the

weekdays:

# Normalize to sum to $1$

In: party$\_$pcts $=$ party$\_$counts.div$($party$\_$counts.sum$(1),$ axis$=0)$

In: party$\_$pcts

size $2345$

day

Fri $0\mathrm{.}8888890\mathrm{.}0555560\mathrm{.}0555560\mathrm{.}000000$

Sat $0\mathrm{.}6235290\mathrm{.}2117650\mathrm{.}1529410\mathrm{.}011765$

Sun $0\mathrm{.}5200000\mathrm{.}2000000\mathrm{.}2400000\mathrm{.}040000$

Thu $0\mathrm{.}8275860\mathrm{.}0689660\mathrm{.}0862070\mathrm{.}017241$

$6.$ Draw plot$($kind$=$'bar'$)$ of the party$\_$pcts from step $5.$

Show your plot output.

$7.$ To find the percentage of the tip for each bill, create a column $$tip$\_$pct$$:

In: tips$[\text{'}$tip$\_$pct$\text{'}]=$ tips$[\text{'}$tip$\text{'}]/$ tips$[\text{'}$totbill$\text{'}]$

tips.head$(5)$

Similarly, to list the first five rows of the dataset:

tips$[$:$6]$

Show your output.

$8.$ Draw a histogram plot $($frequency of the data points split into discrete, evenly spaced bins, with

the number of data points in each bin$),$ using the tip percentages of the total bill, tips$[$tip$\_$pct$],$

of step $7$:

tips$[\text{'}$tip$\_$pct$\text{'}].$plot.hist$($bins$=50)$

Show your output.

$9.$ To show a summary statistics of the percentag of tips $$tip$\_$pct$$ for day, smoker, sex, and time

columns, we can call the describe on a groupby object:

tips.groupby$(\text{'}$smoker$\text{'})[\text{'}$tip$\_$pct$\text{'}].$describe$()$

Show your output.

Similarly, show the $$tip$\_$pct$$ aggregated on $$sex$,$day$$ and $$time$$

Show your output.

$10.$ Aggregating a Series or all of the columns of a DataFrame is a matter of using aggregate with the

desired function or calling a method like mean or std$.$ However, we may want to aggregate using

a different function depending on the column, or multiple functions at once.

To group the tips by day and smoker:

grouped $=$ tips.groupby$([\text{'}$day$\text{'},$ 'smoker'$])$

grouped

Out$[130]$:

To check and view the output of the aggregated files:

grouped.describe$()$

Show your output.

$11.$ To calculate the average of $$tip$\_$pct$$ of the aggregated group $$grouped$\_$pct$$ on $$day$$ and

$$smoker$$

grouped$\_$pct $=$ grouped$[\text{'}$tip$\_$pct$\text{'}]$

grouped$\_$pct$.$agg$(\text{'}$mean$\text{'})$

day smoker

Fri No $0\mathrm{.}151650$

Yes $0\mathrm{.}174783$

Sat No $0\mathrm{.}158048$

Yes $0\mathrm{.}147906$

Sun No $0\mathrm{.}160113$

Yes $0\mathrm{.}187250$

Thu No $0\mathrm{.}160298$

Yes $0\mathrm{.}163863$

Name: tip$\_$pct$,$ dtype: float$64$

$12.$ Alternatively, we can pass a list of functions or function names instead, you get back a

DataFrame with column names taken from the functions:

grouped$\_$pct$.$agg$([\text{'}$mean$\text{'},\text{'}$std$\text{'}])$

mean std

day smoker

Fri No $0\mathrm{.}1516500\mathrm{.}028123$

Yes $0\mathrm{.}1747830\mathrm{.}051293$

Sat No $0\mathrm{.}1580480\mathrm{.}039767$

Yes $0\mathrm{.}1479060\mathrm{.}061375$

Sun No $0\mathrm{.}1601130\mathrm{.}042347$

Yes $0\mathrm{.}1872500\mathrm{.}154134$

Thu No $0\mathrm{.}1602980\mathrm{.}038774$

Yes $0\mathrm{.}1638630\mathrm{.}039389$

$13.$ Similarly, calculate the average of $$tip$\_$pct$$ aggregated on $$day$$ and $$sex$$

Show your code and output.

$14.$ With a DataFrame, we can also specify a list of functions to apply to all of the columns or

different functions per column. To compute the same three statistics for the tip$\_$pct and totalbill

columns:

grouped$[\text{'}$tip$\_$pct$\text{'},$'totbill'$].$agg$([\text{'}$count$\text{'},$'mean', 'max'$])$

tip$\_$pct totbill

count mean max count mean max

day smoker

Fri No $40\mathrm{.}1516500\mathrm{.}187735418\mathrm{.}42000022\mathrm{.}75$

Yes $150\mathrm{.}1747830\mathrm{.}2634801516\mathrm{.}81333340\mathrm{.}17$

Sat No $450\mathrm{.}1580480\mathrm{.}2919904519\mathrm{.}66177848\mathrm{.}33$

Yes $420\mathrm{.}1479060\mathrm{.}3257334221\mathrm{.}27666750\mathrm{.}81$

Sun No $570\mathrm{.}1601130\mathrm{.}2526725720\mathrm{.}50666748\mathrm{.}17$

Yes $190\mathrm{.}1872500\mathrm{.}7103451924\mathrm{.}12000045\mathrm{.}35$

Thu No $450\mathrm{.}1602980\mathrm{.}2663124517\mathrm{.}11311141\mathrm{.}19$

Yes $170\mathrm{.}1638630\mathrm{.}2412551719\mathrm{.}19058843\mathrm{.}11$

$15.$ Suppose we wanted to select the top five tip$\_$pct values by group. First, to write a function that

selects the rows with the largest values in a particular column:

In: def top$($df$,$ n$=5,$ column$=$'tip$\_$pct$\text{'})$: return df$.$sort$()[-$n:$]$

top$($tips$,$ n$=5)$

Show your output.

then, to group by smoker and call apply with the top$()$ function: