RahulVarma_LAB5
pdf
keyboard_arrow_up
School
University of North Texas *
*We aren’t endorsed by this school
Course
5709
Subject
Statistics
Date
Feb 20, 2024
Type
Pages
18
Uploaded by venkatasai1999
3/9/23, 11:32 AM
RahulVarma_LAB5 (2) - Jupyter Notebook
localhost:8889/notebooks/Downloads/RahulVarma_LAB5 (2).ipynb#
1/18
Part 1: Data Visualizations
1. Grouped Bar Plots:
Make both a side-by-side bar plot and a stacked bar plot that displays the number of child
visitors and the number of adult visitors at a waterpark in the months of April, May, June and
July. Be sure to include titles, legends and appropriate labels sufficiently sized for readability.
April
Children: 780 Adults: 315
May
Children: 1050 Adults: 400
June
Children: 3056 Adults: 1000
July
Children: 5025 Adults: 1500
In [93]:
import
numpy as
np
import
matplotlib.pyplot as
plt
3/9/23, 11:32 AM
RahulVarma_LAB5 (2) - Jupyter Notebook
localhost:8889/notebooks/Downloads/RahulVarma_LAB5 (2).ipynb#
2/18
In [164]:
2. Histogram:
Make a histogram of the following scores from the Fall 2017 Data Structures course at Loyola
University Chicago. Feel free to experiment on the best number of histogram bins for
visualization. 114.8, 98.8, 97.3, 96, 94.1, 93.1, 93.1, 91.6, 91.5, 91.3, 90.3, 89.2, 87.5, 87.4,
85.2, 81.7, 81.6, 81.5, 80, 79.3, 78.2, 77.6, 77.1, 76.7, 75.1, 73.9, 72, 71, 64.6, 63.3, 47.2,
38.7
Out[164]:
<matplotlib.legend.Legend at 0x1f2e1d20e80>
months =
[
"April"
, "May"
, "June"
, "July"
]
children =
[
780
, 1050
, 3056
, 5025
]
adults =
[
315
, 400
, 1000
, 1500
]
#code for graph
width =
0.4
bars =
np.arange(
len
(months))
plt.bar(bars
-
0.2
, children, width, label =
"children"
)
plt.bar(bars
+
0.2
, adults, width, label =
"adults"
)
plt.xticks(bars,months)
plt.xlabel(
"Months"
)
plt.ylabel(
" No of visitors"
)
plt.title(
"Number of child visitors vs Number of adult visitors"
)
plt.legend()
3/9/23, 11:32 AM
RahulVarma_LAB5 (2) - Jupyter Notebook
localhost:8889/notebooks/Downloads/RahulVarma_LAB5 (2).ipynb#
3/18
In [96]:
In [97]:
In [98]:
In [165]:
38.7
114.8
Out[165]:
Text(0.5, 1.0, 'Fall 2017- Data Structures Course Scrores ')
FallScores_2017 =
np.array([
114.8
, 98.8
, 97.3
, 96
, 94.1
, 93.1
, 93.1
, 91.6
, 91
])
print
(
min
(FallScores_2017))
print
(
max
(FallScores_2017))
t =
[
0
,
10
,
20
,
30
,
40
,
50
,
60
,
70
,
80
,
90
,
100
,
110
,
120
]
plt.hist(FallScores_2017, bins =
t , ec =
"black" , color =
'#4e7eb5' )
plt.xticks(t)
plt.xlabel(
"scores"
)
plt.ylabel(
"count"
)
plt.title(
"Fall 2017- Data Structures Course Scrores "
)
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
3/9/23, 11:32 AM
RahulVarma_LAB5 (2) - Jupyter Notebook
localhost:8889/notebooks/Downloads/RahulVarma_LAB5 (2).ipynb#
4/18
3. Line Plot:
Create a line plot of sin(x) and cos(x + π/2) for -2π < x < 2π where x increases at intervals of
π/4. 1)Make the sin(x) graph red and make the cos(x+π/2) graph green a)Put both lines onto
the same plot
In [100]:
2)Using the same info as above, make a subplot with 2 different graphs
a)one graph for sin(x) and
Out[100]:
<matplotlib.legend.Legend at 0x1f2dbfb96a0>
x =
np.arange(
-
2
*
np.pi,
2
*
np.pi,np.pi
/
4
) y =
np.sin(x)
z =
np.cos(x +
np.pi
/
2
)
plt.plot(x,y,color
=
'blue'
,label =
"sin(x)"
)
plt.plot(x,z,color
=
'green'
, label =
"cos(x+π/2)"
)
plt.xlabel(
'-2π to 2π'
) plt.ylabel(
'sin(x) and cos(x+ π/2)'
)
plt.title(
'sin(x) and cos(x+ π/2) line plot'
)
plt.legend()
3/9/23, 11:32 AM
RahulVarma_LAB5 (2) - Jupyter Notebook
localhost:8889/notebooks/Downloads/RahulVarma_LAB5 (2).ipynb#
5/18
In [101]:
b)one graph for cos(x+π/2)
Out[101]:
<matplotlib.legend.Legend at 0x1f2dc0ab0d0>
x =
np.arange(
-
2
*
np.pi,
2
*
np.pi,np.pi
/
4
)
y =
np.sin(x)
plt.plot(x,y,color
=
'green'
, label =
"sin(x)"
)
plt.xlabel(
' -2π to 2π'
) plt.ylabel(
'sin(x)'
)
plt.title(
'sin(x) line plot, ranging from -2π to 2π'
)
plt.legend()
3/9/23, 11:32 AM
RahulVarma_LAB5 (2) - Jupyter Notebook
localhost:8889/notebooks/Downloads/RahulVarma_LAB5 (2).ipynb#
6/18
In [166]:
Using the following data about winter temperatures affecting the number of days for lake ice at
Lake Superior, construct a scatter plot to display the data. Include a line of best fit.
Mean Temperature (in Fahrenheit): 22.94, 23.02, 25.68, 19.96, 24.80, 23.98, 22.10, 20.30,
24.20, 22.74, 24.16, 24.94, 22.40, 22.14, 20.84, 25.66, 21.73, 24.49, 24.13, 22.17, 21.73,
20.41, 24.41, 23.95, 20.95, 26.71, 22.81, 23.11, 23.33, 28.83, 23.11, 21.47, 23.97, 24.75,
23.61, 23.08, 21.24, 26.63, 23.88 Days of Ice: 87, 137, 106, 97, 105, 118, 118, 136, 91, 107,
96, 114, 125, 115, 118, 82, 115, 97, 104, 146, 126, 141, 111, 123, 118, 83, 48, 118, 116, 81,
116, 123, 112, 99, 102, 118, 63, 62, 132
In [103]:
Out[166]:
<matplotlib.legend.Legend at 0x1f2e1e28220>
x =
np.arange(
-
2
*
np.pi,
2
*
np.pi,np.pi
/
4
) y =
np.cos(x +
np.pi
/
2
)
plt.plot(x,z,color
=
'red'
, label =
"cos(x+π/2)"
)
plt.xlabel(
'-2π to 2π'
) plt.ylabel(
'cos(x+ π/2)'
)
plt.title(
'line plot, ranging from -2π to 2π'
)
plt.legend()
Mean_Temperature =
np.array([
22.94
, 23.02
, 25.68
, 19.96
, 24.80
, 23.98
, 22.10
, Days_of_Ice =
np.array([
87
, 137
, 106
, 97
, 105
, 118
, 118
, 136
, 91
, 107
, 96
, 114
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
3/9/23, 11:32 AM
RahulVarma_LAB5 (2) - Jupyter Notebook
localhost:8889/notebooks/Downloads/RahulVarma_LAB5 (2).ipynb#
7/18
In [145]:
Part 2: Basic Data Structure
1: Lists
1) Make a list with the spelled-out number strings ‘one’, ‘two’, ‘three’, ‘four’, and ‘five’ in that
order and call it myList.
In [2]:
2) Remove ‘three’ from the list using positional indexing.
Out[145]:
[Text(0.5, 0, 'Mean temperature'),
Text(0, 0.5, 'Days of ice'),
Text(0.5, 1.0, 'Mean Temperature vs Days of Ice')]
import
seaborn as
sns
plot =
sns.regplot(Mean_Temperature,Days_of_Ice)
plot.set(xlabel =
"Mean temperature"
, ylabel =
"Days of ice"
, title =
"Mean Te
mylist =
[
"one"
, "two"
, "three"
, "four"
, "five"
]
3/9/23, 11:32 AM
RahulVarma_LAB5 (2) - Jupyter Notebook
localhost:8889/notebooks/Downloads/RahulVarma_LAB5 (2).ipynb#
8/18
In [3]:
3) Check if ‘four’ is in the list.
In [4]:
4) Append ‘six’ to the end of the list, then print the length of the list.
In [5]:
In [6]:
5) Print the contents of the list, but also next to each item print the length of the string (e.g. one
is 3, four is 4) using a for loop.
In [109]:
6) Create a list only of the lengths of the strings and show your result. You can use the loop
before to fill the list.
In [167]:
Out[3]:
['one', 'two', 'four', 'five']
True
Out[5]:
['one', 'two', 'four', 'five', 'six']
Out[6]:
5
one is 3
two is 3
four is 4
five is 4
six is 3
Out[167]:
[3, 3, 4, 4, 3]
mylist
del
mylist[
2
]
mylist
print (
"four" in
mylist)
mylist.append(
"six"
)
mylist
len
(mylist)
for
i in
mylist:
print
(i ,
"is"
,
len
(i))
length_string
=
[
len
(word) for
word in
mylist]
length_string
3/9/23, 11:32 AM
RahulVarma_LAB5 (2) - Jupyter Notebook
localhost:8889/notebooks/Downloads/RahulVarma_LAB5 (2).ipynb#
9/18
2: Dictionaries
1) Make a dictionary with the keys be English words as below, and the values be the
translation. You can use this language example (German) or choose your own. Note: you need
to make sure all of these words are represented as strings, in quotes. apple - Apfel apples -
Äpfel I - Ich and - und like - mag strawberries - Erdbeeren
In [168]:
2) Use the dictionary to look up the translation for ‘apple’ and ‘like’.
In [169]:
3) Make a variable var with the string “I like apples and strawberries”.
In [173]:
4) Now create a list from var with each word a separate item (this is a string split operation).
In [174]:
In [175]:
5) Iterate through the list you’ve created and replace any word in your dictionary with the
translation.
Out[168]:
{'apple': 'Apfel',
'apples': 'Äpfel',
'I': 'Ich',
'and': 'und',
'like': 'mag',
'strawberries': 'Erdbeeren'}
Apfel
mag
Out[173]:
'I like apples and strawberries'
Out[175]:
['I', 'like', 'apples', 'and', 'strawberries']
dictionary_words =
{
"apple"
:
"Apfel"
, "apples"
:
"Äpfel"
, "I"
:
"Ich"
, "and"
:
"und"
dictionary_words
print (dictionary_words[
"apple"
])
print (dictionary_words[
"like"
])
var =
"I like apples and strawberries"
var
split_string =
var.split()
split_string
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
3/9/23, 11:32 AM
RahulVarma_LAB5 (2) - Jupyter Notebook
localhost:8889/notebooks/Downloads/RahulVarma_LAB5 (2).ipynb#
10/18
In [177]:
6) Now take your new list and turn it into a string with spaces between the words.
In [178]:
3: Arrays
1) Create an array of zeros of size 8 x 8 and print the data type of the array.
In [179]:
In [180]:
2) Fill the array with the numbers 1 to 64 first by row, then by column. You may want to use a
for loop inside a for loop to do this.
Out[177]:
['Ich', 'mag', 'Äpfel', 'und', 'Erdbeeren']
Out[178]:
'Ich mag Äpfel und Erdbeeren'
Out[179]:
array([[0., 0., 0., 0., 0., 0., 0., 0.],
[0., 0., 0., 0., 0., 0., 0., 0.],
[0., 0., 0., 0., 0., 0., 0., 0.],
[0., 0., 0., 0., 0., 0., 0., 0.],
[0., 0., 0., 0., 0., 0., 0., 0.],
[0., 0., 0., 0., 0., 0., 0., 0.],
[0., 0., 0., 0., 0., 0., 0., 0.],
[0., 0., 0., 0., 0., 0., 0., 0.]])
Out[180]:
numpy.ndarray
replace_list =
[]
for
item in
split_string:
if
item in
dictionary_words:
replace_list
.append(dictionary_words[item])
else
:
replace_list.append(item)
replace_list
replace_string =
' '
.join(replace_list)
replace_string
array =
np.zeros((
8
,
8
))
array
type
(array)
3/9/23, 11:32 AM
RahulVarma_LAB5 (2) - Jupyter Notebook
localhost:8889/notebooks/Downloads/RahulVarma_LAB5 (2).ipynb#
11/18
In [182]:
3) Transpose the array.
In [183]:
4) Print only the top 4 rows and columns.
In [184]:
5) Make a 1D array out of your 2D array with the numbers 1 to 64 in order (note the column vs
row issue, you may need transposes.)
In [187]:
Out[182]:
array([[ 1, 2, 3, 4, 5, 6, 7, 8],
[ 9, 10, 11, 12, 13, 14, 15, 16],
[17, 18, 19, 20, 21, 22, 23, 24],
[25, 26, 27, 28, 29, 30, 31, 32],
[33, 34, 35, 36, 37, 38, 39, 40],
[41, 42, 43, 44, 45, 46, 47, 48],
[49, 50, 51, 52, 53, 54, 55, 56],
[57, 58, 59, 60, 61, 62, 63, 64]])
Out[183]:
array([[ 1, 9, 17, 25, 33, 41, 49, 57],
[ 2, 10, 18, 26, 34, 42, 50, 58],
[ 3, 11, 19, 27, 35, 43, 51, 59],
[ 4, 12, 20, 28, 36, 44, 52, 60],
[ 5, 13, 21, 29, 37, 45, 53, 61],
[ 6, 14, 22, 30, 38, 46, 54, 62],
[ 7, 15, 23, 31, 39, 47, 55, 63],
[ 8, 16, 24, 32, 40, 48, 56, 64]])
Out[184]:
array([[ 1, 9, 17, 25],
[ 2, 10, 18, 26],
[ 3, 11, 19, 27],
[ 4, 12, 20, 28]])
Out[187]:
array([ 1, 9, 17, 25, 33, 41, 49, 57, 2, 10, 18, 26, 34, 42, 50, 58, 3,
11, 19, 27, 35, 43, 51, 59, 4, 12, 20, 28, 36, 44, 52, 60, 5, 13,
21, 29, 37, 45, 53, 61, 6, 14, 22, 30, 38, 46, 54, 62, 7, 15, 23,
31, 39, 47, 55, 63, 8, 16, 24, 32, 40, 48, 56, 64])
s =
len
(array)
k =
len
(array)
fill_array =
[[(j
+
1
)
+
s
*
i for
j in
range
(s)] for
i in
range
(k)]
fill_array
=
np.array(fill_array)
fill_array
array =
np.transpose(fill_array)
array
array[
0
:
4
,
0
:
4
]
array_1 =
array.flatten()
array_1
3/9/23, 11:32 AM
RahulVarma_LAB5 (2) - Jupyter Notebook
localhost:8889/notebooks/Downloads/RahulVarma_LAB5 (2).ipynb#
12/18
In [189]:
6) Now take that 1D array you made from before and reshape it back to the original 2D array.
In [193]:
In [194]:
Part 3: Data Frames
In this part, we will study a classic data set - the survivors in the sinking of the Titanic. As there
were limited lifeboats, decisions were made prioritizing who would and would not survive. We
will observe how different factors such as age, sex, and class affected a person’s chance of
survival using data frames.
Steps:
1. Input the following data into a data frame called titanic, and display the entire data frame:
Sex, Class, Survived, Died Children, First, 6, 0 Children, Second, 24, 0 Children, Third,
27, 52 Men, First, 57, 118 Men, Second, 14, 154 Men, Third, 75, 387 Men, Crew, 192, 693
Women, First, 140, 4 Women, Second, 80, 13 Women, Third, 76, 89 Women, Crew, 20, 3
data.
In [127]:
In [196]:
Out[189]:
1
Out[193]:
array([[ 1, 9, 17, 25, 33, 41, 49, 57],
[ 2, 10, 18, 26, 34, 42, 50, 58],
[ 3, 11, 19, 27, 35, 43, 51, 59],
[ 4, 12, 20, 28, 36, 44, 52, 60],
[ 5, 13, 21, 29, 37, 45, 53, 61],
[ 6, 14, 22, 30, 38, 46, 54, 62],
[ 7, 15, 23, 31, 39, 47, 55, 63],
[ 8, 16, 24, 32, 40, 48, 56, 64]])
Out[194]:
2
array_1.ndim
array_2 =
array_1.reshape(
8
,
8
)
array_2
array_2.ndim
import
pandas as
pd
titanic_data =
{
"sex"
:[
"Children"
, "Children"
, "Children"
, "Men"
, "Men"
, "Men
"Class"
:[
"First"
, "Second"
, "Third"
, "First"
, "Second"
, "Third"
, "Crew
"Survived"
:[
6
,
24
,
27
,
57
,
14
,
75
,
192
,
140
,
80
,
76
,
20
],
"Died"
:[
0
,
0
,
52
,
118
,
154
,
387
,
693
,
4
,
13
,
89
,
3
]}
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
3/9/23, 11:32 AM
RahulVarma_LAB5 (2) - Jupyter Notebook
localhost:8889/notebooks/Downloads/RahulVarma_LAB5 (2).ipynb#
13/18
In [197]:
2. Now only show the data of the people in first class.
In [198]:
3. Delete the crew members from the data.
Out[197]:
sex
Class
Survived
Died
0
Children
First
6
0
1
Children
Second
24
0
2
Children
Third
27
52
3
Men
First
57
118
4
Men
Second
14
154
5
Men
Third
75
387
6
Men
Crew
192
693
7
Women
First
140
4
8
Women
Second
80
13
9
Women
Third
76
89
10
women
Crew
20
3
Out[198]:
sex
Class
Survived
Died
0
Children
First
6
0
3
Men
First
57
118
7
Women
First
140
4
data_frame =
pd.DataFrame(data
=
titanic_data, columns =
[
"sex"
, "Class"
, "Survi
data_frame
data_frame.loc[data_frame[
"Class"
] ==
"First"
]
3/9/23, 11:32 AM
RahulVarma_LAB5 (2) - Jupyter Notebook
localhost:8889/notebooks/Downloads/RahulVarma_LAB5 (2).ipynb#
14/18
In [199]:
4. Create a new column that is the total number of people for that group (those who survived
+ died).
In [200]:
5. Create a new column with the percentage of people who survived.
Out[199]:
sex
Class
Survived
Died
0
Children
First
6
0
1
Children
Second
24
0
2
Children
Third
27
52
3
Men
First
57
118
4
Men
Second
14
154
5
Men
Third
75
387
7
Women
First
140
4
8
Women
Second
80
13
9
Women
Third
76
89
Out[200]:
sex
Class
Survived
Died
Total_Number
0
Children
First
6
0
6
1
Children
Second
24
0
24
2
Children
Third
27
52
79
3
Men
First
57
118
175
4
Men
Second
14
154
168
5
Men
Third
75
387
462
7
Women
First
140
4
144
8
Women
Second
80
13
93
9
Women
Third
76
89
165
data_frame.drop(labels =
[
6
,
10
], inplace =
True
)
data_frame
data_frame[
"Total_Number"
] =
data_frame[
"Survived"
]
+
data_frame[
"Died"
]
data_frame
3/9/23, 11:32 AM
RahulVarma_LAB5 (2) - Jupyter Notebook
localhost:8889/notebooks/Downloads/RahulVarma_LAB5 (2).ipynb#
15/18
In [201]:
6. Delete the column indicating the total number of people in that group
In [202]:
7. Only show the rows where more than 80% of the people survived.
Out[201]:
sex
Class
Survived
Died
Total_Number
percentage_survived
0
Children
First
6
0
6
100.0
1
Children
Second
24
0
24
100.0
2
Children
Third
27
52
79
34.0
3
Men
First
57
118
175
33.0
4
Men
Second
14
154
168
8.0
5
Men
Third
75
387
462
16.0
7
Women
First
140
4
144
97.0
8
Women
Second
80
13
93
86.0
9
Women
Third
76
89
165
46.0
Out[202]:
sex
Class
Survived
Died
percentage_survived
0
Children
First
6
0
100.0
1
Children
Second
24
0
100.0
2
Children
Third
27
52
34.0
3
Men
First
57
118
33.0
4
Men
Second
14
154
8.0
5
Men
Third
75
387
16.0
7
Women
First
140
4
97.0
8
Women
Second
80
13
86.0
9
Women
Third
76
89
46.0
data_frame[
"percentage_survived"
] =
round
((data_frame[
"Survived"
]
/
data_frame[
data_frame
data_frame.drop(labels =
"Total_Number"
, axis =
1
, inplace =
True
)
data_frame
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
3/9/23, 11:32 AM
RahulVarma_LAB5 (2) - Jupyter Notebook
localhost:8889/notebooks/Downloads/RahulVarma_LAB5 (2).ipynb#
16/18
In [203]:
8. Then only show the rows where less than 40% of the people survived.
In [204]:
9. Calculate the total number of people that survived and died for each class, then report the
percentages. (Hint: Use a grouped calculation.)
In [205]:
Out[203]:
sex
Class
Survived
Died
percentage_survived
0
Children
First
6
0
100.0
1
Children
Second
24
0
100.0
7
Women
First
140
4
97.0
8
Women
Second
80
13
86.0
Out[204]:
sex
Class
Survived
Died
percentage_survived
2
Children
Third
27
52
34.0
3
Men
First
57
118
33.0
4
Men
Second
14
154
8.0
5
Men
Third
75
387
16.0
Out[205]:
sex
Class
Survived
Died
0
Children
First
6
0
1
Children
Second
24
0
2
Children
Third
27
52
3
Men
First
57
118
4
Men
Second
14
154
5
Men
Third
75
387
7
Women
First
140
4
8
Women
Second
80
13
9
Women
Third
76
89
data_frame[data_frame[
"percentage_survived"
]
>
80
]
data_frame[data_frame[
"percentage_survived"
]
<
40
]
data_frame.drop(columns =
"percentage_survived"
, axis =
1
, inplace =
True
)
data_frame
3/9/23, 11:32 AM
RahulVarma_LAB5 (2) - Jupyter Notebook
localhost:8889/notebooks/Downloads/RahulVarma_LAB5 (2).ipynb#
17/18
In [206]:
In [207]:
In [208]:
10. Save your table in CSV format (as e.g. titanic_data.csv) with the first line as headers for
the columns.
Out[206]:
sex
Class
Survived
Died
total
0
Children
First
6
0
6
1
Children
Second
24
0
24
2
Children
Third
27
52
79
3
Men
First
57
118
175
4
Men
Second
14
154
168
5
Men
Third
75
387
462
7
Women
First
140
4
144
8
Women
Second
80
13
93
9
Women
Third
76
89
165
Class
First 62.0
Second 41.0
Third 25.0
dtype: float64
Class
First 38.0
Second 59.0
Third 75.0
dtype: float64
Out[208]:
% survived
% died
Class
First
62.0
38.0
Second
41.0
59.0
Third
25.0
75.0
data_frame[
"total"
] =
df[
"Survived"
] +
df[
"Died"
]
data_frame
percentage_survived =
round
((data_frame.groupby(
"Class"
).Survived.sum()
/
data_f
percentage_died =
round
((data_frame.groupby(
"Class"
).Died.sum()
/
data_frame.gro
print (percentage_survived)
print (percentage_died)
results =
pd.concat([percentage_survived, percentage_died], axis
=
1
, keys
=
[
"% results
3/9/23, 11:32 AM
RahulVarma_LAB5 (2) - Jupyter Notebook
localhost:8889/notebooks/Downloads/RahulVarma_LAB5 (2).ipynb#
18/18
In [209]:
11. Duplicate the CSV file on your computer since you will be editing the copied version (e.g.
titanic_data2.csv). Open the new CSV file in a text editor. Note the way the data is
organized. Now, in the text editor, add new lines including the data for the crew that was
removed earlier. (Help: the percentage of male crew and female crew that survived was
21.69% and 86.96%.)
In [210]:
12. Now read that updated CSV file into a new data frame called titanic2, and display the
data.
In [212]:
Out[212]:
Class
% survived
% died
0
First
62.0
38.0
1
Second
41.0
59.0
2
Third
25.0
75.0
results.to_csv(
"titanic_data.csv"
)
results.to_csv(
"titanic_data2.csv"
)
titanic2 =
pd.read_csv(
"titanic_data2.csv"
)
titanic2
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
Recommended textbooks for you
![Text book image](https://www.bartleby.com/isbn_cover_images/9780079039897/9780079039897_smallCoverImage.jpg)
Glencoe Algebra 1, Student Edition, 9780079039897...
Algebra
ISBN:9780079039897
Author:Carter
Publisher:McGraw Hill
![Text book image](https://www.bartleby.com/isbn_cover_images/9781680331141/9781680331141_smallCoverImage.jpg)
Big Ideas Math A Bridge To Success Algebra 1: Stu...
Algebra
ISBN:9781680331141
Author:HOUGHTON MIFFLIN HARCOURT
Publisher:Houghton Mifflin Harcourt
Algebra & Trigonometry with Analytic Geometry
Algebra
ISBN:9781133382119
Author:Swokowski
Publisher:Cengage
![Text book image](https://www.bartleby.com/isbn_cover_images/9781938168383/9781938168383_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9780547587776/9780547587776_smallCoverImage.jpg)
Holt Mcdougal Larson Pre-algebra: Student Edition...
Algebra
ISBN:9780547587776
Author:HOLT MCDOUGAL
Publisher:HOLT MCDOUGAL
Recommended textbooks for you
- Glencoe Algebra 1, Student Edition, 9780079039897...AlgebraISBN:9780079039897Author:CarterPublisher:McGraw HillBig Ideas Math A Bridge To Success Algebra 1: Stu...AlgebraISBN:9781680331141Author:HOUGHTON MIFFLIN HARCOURTPublisher:Houghton Mifflin HarcourtAlgebra & Trigonometry with Analytic GeometryAlgebraISBN:9781133382119Author:SwokowskiPublisher:Cengage
- Holt Mcdougal Larson Pre-algebra: Student Edition...AlgebraISBN:9780547587776Author:HOLT MCDOUGALPublisher:HOLT MCDOUGAL
![Text book image](https://www.bartleby.com/isbn_cover_images/9780079039897/9780079039897_smallCoverImage.jpg)
Glencoe Algebra 1, Student Edition, 9780079039897...
Algebra
ISBN:9780079039897
Author:Carter
Publisher:McGraw Hill
![Text book image](https://www.bartleby.com/isbn_cover_images/9781680331141/9781680331141_smallCoverImage.jpg)
Big Ideas Math A Bridge To Success Algebra 1: Stu...
Algebra
ISBN:9781680331141
Author:HOUGHTON MIFFLIN HARCOURT
Publisher:Houghton Mifflin Harcourt
Algebra & Trigonometry with Analytic Geometry
Algebra
ISBN:9781133382119
Author:Swokowski
Publisher:Cengage
![Text book image](https://www.bartleby.com/isbn_cover_images/9781938168383/9781938168383_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9780547587776/9780547587776_smallCoverImage.jpg)
Holt Mcdougal Larson Pre-algebra: Student Edition...
Algebra
ISBN:9780547587776
Author:HOLT MCDOUGAL
Publisher:HOLT MCDOUGAL