Week 3 Tuesday Discussion¶

Quiz 2 Review¶

import seaborn as sns

df = sns.load_dataset("titanic")
df

	survived	pclass	sex	age	sibsp	parch	fare	embarked	class	who	adult_male	deck	embark_town	alive	alone
0	0	3	male	22.0	1	0	7.2500	S	Third	man	True	NaN	Southampton	no	False
1	1	1	female	38.0	1	0	71.2833	C	First	woman	False	C	Cherbourg	yes	False
2	1	3	female	26.0	0	0	7.9250	S	Third	woman	False	NaN	Southampton	yes	True
3	1	1	female	35.0	1	0	53.1000	S	First	woman	False	C	Southampton	yes	False
4	0	3	male	35.0	0	0	8.0500	S	Third	man	True	NaN	Southampton	no	True
...	...	...	...	...	...	...	...	...	...	...	...	...	...	...	...
886	0	2	male	27.0	0	0	13.0000	S	Second	man	True	NaN	Southampton	no	True
887	1	1	female	19.0	0	0	30.0000	S	First	woman	False	B	Southampton	yes	True
888	0	3	female	NaN	1	2	23.4500	S	Third	woman	False	NaN	Southampton	no	False
889	1	1	male	26.0	0	0	30.0000	C	First	man	True	C	Cherbourg	yes	True
890	0	3	male	32.0	0	0	7.7500	Q	Third	man	True	NaN	Queenstown	no	True

891 rows × 15 columns

Question 1:

Write code to determine which columns of df contain missing values. Do the same to determine which rows of df contain missing values.
Write code to determine which rows of df contain at least 2 missing values.
Create a sub-DataFrame of df which contains only rows which have at least one missing value.
Create a sub-DataFrame of df which contains only rows which have at least one missing value and embark_town listed as Southampton.

Code-Along Solutions¶

[c for c in df.columns if df[c].isna().any()]

['age', 'embarked', 'deck', 'embark_town']

[r for r in df.index if df.iloc[r].isna().any()]

[r for r in df.index if df.iloc[r].isna().sum() >= 2]

df.loc[[r for r in df.index if df.loc[r].isna().any()]]

	survived	pclass	sex	age	sibsp	parch	fare	embarked	class	who	adult_male	deck	embark_town	alive	alone
0	0	3	male	22.0	1	0	7.2500	S	Third	man	True	NaN	Southampton	no	False
2	1	3	female	26.0	0	0	7.9250	S	Third	woman	False	NaN	Southampton	yes	True
4	0	3	male	35.0	0	0	8.0500	S	Third	man	True	NaN	Southampton	no	True
5	0	3	male	NaN	0	0	8.4583	Q	Third	man	True	NaN	Queenstown	no	True
7	0	3	male	2.0	3	1	21.0750	S	Third	child	False	NaN	Southampton	no	False
...	...	...	...	...	...	...	...	...	...	...	...	...	...	...	...
884	0	3	male	25.0	0	0	7.0500	S	Third	man	True	NaN	Southampton	no	True
885	0	3	female	39.0	0	5	29.1250	Q	Third	woman	False	NaN	Queenstown	no	False
886	0	2	male	27.0	0	0	13.0000	S	Second	man	True	NaN	Southampton	no	True
888	0	3	female	NaN	1	2	23.4500	S	Third	woman	False	NaN	Southampton	no	False
890	0	3	male	32.0	0	0	7.7500	Q	Third	man	True	NaN	Queenstown	no	True

709 rows × 15 columns

Pre-Written Solutions¶

#Missing columns
[c for c in df.columns if df[c].isna().any()]

['age', 'embarked', 'deck', 'embark_town']

#Missing rows
[r for r in df.index if df.loc[r].isna().any()]

#At least 2 missing rows
more_1 = [r for r in df.index if df.loc[r].isna().sum() > 1]
df.loc[more_1].isna().sum(axis = 1)

    2
   2
   2
   2
   2
      ..
  2
  2
  2
  2
  2
Length: 160, dtype: int64

#Missing sub-DataFrame
df.loc[[r for r in df.index if df.loc[r].isna().any()]]

	survived	pclass	sex	age	sibsp	parch	fare	embarked	class	who	adult_male	deck	embark_town	alive	alone
0	0	3	male	22.0	1	0	7.2500	S	Third	man	True	NaN	Southampton	no	False
2	1	3	female	26.0	0	0	7.9250	S	Third	woman	False	NaN	Southampton	yes	True
4	0	3	male	35.0	0	0	8.0500	S	Third	man	True	NaN	Southampton	no	True
5	0	3	male	NaN	0	0	8.4583	Q	Third	man	True	NaN	Queenstown	no	True
7	0	3	male	2.0	3	1	21.0750	S	Third	child	False	NaN	Southampton	no	False
...	...	...	...	...	...	...	...	...	...	...	...	...	...	...	...
884	0	3	male	25.0	0	0	7.0500	S	Third	man	True	NaN	Southampton	no	True
885	0	3	female	39.0	0	5	29.1250	Q	Third	woman	False	NaN	Queenstown	no	False
886	0	2	male	27.0	0	0	13.0000	S	Second	man	True	NaN	Southampton	no	True
888	0	3	female	NaN	1	2	23.4500	S	Third	woman	False	NaN	Southampton	no	False
890	0	3	male	32.0	0	0	7.7500	Q	Third	man	True	NaN	Queenstown	no	True

709 rows × 15 columns

#Missing + embark_town
df.loc[[r for r in df.index if df.loc[r].isna().any() and df.loc[r,"embark_town"] == "Southampton"]]

	survived	pclass	sex	age	sibsp	parch	fare	embarked	class	who	adult_male	deck	embark_town	alive	alone
0	0	3	male	22.0	1	0	7.2500	S	Third	man	True	NaN	Southampton	no	False
2	1	3	female	26.0	0	0	7.9250	S	Third	woman	False	NaN	Southampton	yes	True
4	0	3	male	35.0	0	0	8.0500	S	Third	man	True	NaN	Southampton	no	True
7	0	3	male	2.0	3	1	21.0750	S	Third	child	False	NaN	Southampton	no	False
8	1	3	female	27.0	0	2	11.1333	S	Third	woman	False	NaN	Southampton	yes	False
...	...	...	...	...	...	...	...	...	...	...	...	...	...	...	...
882	0	3	female	22.0	0	0	10.5167	S	Third	woman	False	NaN	Southampton	no	True
883	0	2	male	28.0	0	0	10.5000	S	Second	man	True	NaN	Southampton	no	True
884	0	3	male	25.0	0	0	7.0500	S	Third	man	True	NaN	Southampton	no	True
886	0	2	male	27.0	0	0	13.0000	S	Second	man	True	NaN	Southampton	no	True
888	0	3	female	NaN	1	2	23.4500	S	Third	woman	False	NaN	Southampton	no	False

529 rows × 15 columns

Question 2:

Consider the following Altair chart. Explain how you could use it to determine where parch is the largest and how many different unique embark_town values there are.

import altair as alt
alt.Chart(df[df["age"] < 20]).mark_circle().encode(
    x = alt.X("age"),
    y = alt.Y("fare"),
    color = alt.Color("embark_town", legend = None),
    size = alt.Size("parch", legend = None)
)

UC Irvine Math 10 S22

Week 3 Tuesday Discussion

Contents

Week 3 Tuesday Discussion¶

Quiz 2 Review¶

Code-Along Solutions¶

Pre-Written Solutions¶