40 Questions to test your skill in Python for Data Science

A) a.append(b)B) a.extend(b)

C) Any of the above

D) None of these

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import time
str = '21/01/2017'
datetime_value = time.strptime(str,date_format)

6) To convert the above string, what should be written in place of date_format?A) “%d/%m/%y”

B) “%D/%M/%Y”

C) “%d/%M/%y”

D) “%d/%m/%Y”

Solution: (D)

Option D is correct

Note: Library numpy has been imported as np.A) np.eye(3)

B) identity(3)

C) np.array([1, 0, 0], [0, 1, 0], [0, 0, 1])

D) All of these

Solution: (A)
Option B does not exist (it should be np.identity()). And option C is wrong, because the syntax is incorrect. So the answer is option A

I have two numpy arrays “e” and “f”.You get the following output when you print “e” & “f”

print e
[1, 2, 3, 2, 3, 4, 4, 5, 6]
print f
[[1, 2, 3], [2, 3, 4], [4, 5, 6]]

When you change the values of the first array, the values for the second array also changes. This creates a problem while processing the data.

For example, if you set the first 5 values of e as 0; i.e.

print e[:5]

0

the final values of e and f are

print e 

[0, 0, 0, 0, 0, 4, 4, 5, 6]

print f

[[0, 0, 0], [0, 0, 4], [4, 5, 6]]

You surmise that the two arrays must have the same space allocated.

A) Check memory of both arrays, if they match that means the arrays are same.

B) Do “np.array_equal(e, f)” and if the output is “True” then they both are same

C) Print flags of both arrays by e.flags and f.flags; check the flag “OWNDATA”. If one of them is False, then both the arrays have same space allocated.

D) None of these

Solution: (C)

Option C is correct

train_set = np.array([1, 2, 3])
test_set = np.array([[0, 1, 2], [1, 2, 3]])
resulting_set --> [[1, 2, 3], [0, 1, 2], [1, 2, 3]]

9) How would you join the two arrays?

Note: Numpy library has been imported as npA) resulting_set = train_set.append(test_set)

B) resulting_set = np.concatenate([train_set, test_set])

C) resulting_set = np.vstack([train_set, test_set])

D) None of these

Solution: (C)

Both option A and B would do horizontal stacking, but we would like to have vertical stacking. So option C is correct

A) np.random.seed(1)B) np.random.seed(40)

C) np.random.seed(32)

D) Can’t say

Solution: (D)

There is no best value for seed. It depends on the data.

11) Which command will be appropriate to fill missing value while reading the file with numpy? 

Note: numpy has been imported as np

A) filling_values = (“-“, 0, 01/01/2010, 0)
temp = np.genfromtxt(filename, filling_values=filling_values)

B) filling_values = (“-“, 0, 01/01/2010, 0)
temp = np.loadtxt(filename, filling_values=filling_values)

C) filling_values = (“-“, 0, 01/01/2010, 0)
temp = np.gentxt(filename, filling_values=filling_values)

D) None of these

Solution: (A)

Option A is correct

A) from sklearn.decision_tree import DecisionTreeClassifierB) from sklearn.ensemble import DecisionTreeClassifier

C) from sklearn.tree import DecisionTreeClassifier

D) None of these

Solution: (C)

Option C is correct

Note: Library StringIO has been imported as StringIO.

A) 
link = https://docs.google.com/spreadsheets/d/...source = StringIO.StringIO(requests.get(link).content))
data = pd.read_csv(source)

B) 
link = https://docs.google.com/spreadsheets/d/...source = StringIO(request.get(link).content))
data = pd.read_csv(source)

C)

link = https://docs.google.com/spreadsheets/d/...source = StringIO(requests.get(link).content))
data = pd.read_csv(source)

D) None of these

Solution: (A)

Option A is correct

train  = pd.DataFrame({'id':[1,2,4],'features':[["A","B","C"],["A","D","E"],["C","D","F"]]})

Now you want to apply a lambda function on “features” column: 

train['features_t'] = train["features"].apply(lambda x: " ".join(["_".join(i.split(" ")) for i in x]))

14) What will be the output of following print command?

print train['features_t']

A)

0    A B C
1    A D E
2    C D F

B)

0    AB

1    ADE

2    CDF

C) Error
D) None of these

Solution: (A)

Option A is correct

The quality column currently has values 1 to 10, but we want to substitute this by a binary classification problem. You want to keep the threshold for classification to 5, such that if the class is greater than 5, the output should be 1, else output should be 0.

15) Which of the following codes would help you perform this task?

Note: Numpy has been imported as np and dataframe is set as df.A)

Y = df[quality].values
Y = np.array([1 if y >= 6 else 0 for y in Y])

B)

Y = df[quality].values()
Y = np.array([0 if y >= 6 else 1 for y in Y])

C)

Y = df[quality]
Y = np.array([0 if y >= 6 else 1 for y in Y])

D)None of these

Solution: (A)

Option A is correct

Question Context 16

Suppose we make a dataframe as

df = pd.DataFrame(['ff', 'gg', 'hh', 'yy'],
              [24, 12, 48, 30],
               columns = ['Name', 'Age'])

1. df[‘Name’] and
2. df.loc[:, ‘Name’]

Note: Pandas has been imported as pd

A) 1 is view of original dataframe and 2 is a copy of original dataframe.

B) 2 is view of original dataframe and 1 is a copy of original dataframe.

C) Both are copies of original dataframe.

D) Both are views of original dataframe

Solution: (B)

Option B is correct. Refer the official docs of pandas library.

Consider a function “fun” which is defined below:

def fun(x):
    x[0] = 5
    return x

Now you define a list which has three numbers in it.

g = [10,11,12]

A) [5, 11, 12] [5, 11, 12]B) [5, 11, 12] [10, 11, 12]

C) [10, 11, 12] [10, 11, 12]

D) [10, 11, 12] [5, 11, 12]

Solution: (A)

Option A is correct

def sigmoid(x):
   return (1 / (1 + math.exp(-x)))

18) It is necessary to know how to find the derivatives of sigmoid, as it would be essential for backpropagation. Select the option for finding derivative?

A)

import scipy

Dv = scipy.misc.derive(sigmoid)

 B)

from sympy import *

x = symbol(x)

y = sigmoid(x)

Dv = y.differentiate(x)

C)

Dv = sigmoid(x) * (1 - sigmoid(x))

D) None of these

Solution: (C)

Option C is correct

For this, first you have to expand the data for every month (considering that every month has 30 days)

19) Which of the following code would do this?

Note: Numpy has been imported as np and dataframe is set as df.A) new_df = pd.concat([df]*30, index = False)

B) new_df = pd.concat([df]*30, ignore_index=True)

C) new_df = pd.concat([df]*30, ignore_index=False)

D) None of these

Solution: (B)

Option B is correct

df.rename(columns = {'Count':'Click_Count'})

What will be the output of print statement below?

print df.columns

Note: Pandas library has been imported as pd.

A) [‘Click_Id’, ‘Click_Count’]

B) [‘Click_Id’, ‘Count’]

C) Error

D) None of these

Solution: (B)

Option B is correct

df = pd.DataFrame({'Click_Id':['A','B','C','D','E'],'Count':[100,200,300,400,250]})

21) In many data science projects, you are required to convert a dataframe into a dictionary. Suppose you want to convert “df” into a dictionary such that ‘Click_Id’ will be the key and ‘Count’ will be the value for each key. Which of the following options will give you the desired result?

Note: Pandas library has been imported as pdA) set_index(‘Click_Id’)[‘Count’].to_dict()

B) set_index(‘Count’)[‘Click_Id’].to_dict()

C) We cannot perform this task since dataframe and dictionary are different data structures

D) None of these 

Solution: (A)

Option A is correct

df1 = df

Now you want to change some values of “Count” column in df.

df.loc[df.Click_Id == 'A', 'Count'] += 100

Which of the following will be the right output for the below print statement?

print df.Count.values,df1.Count.values

Note: Pandas library has been imported as pd.

A) [200 200 300 400 250] [200 200 300 400 250]

B) [100 200 300 400 250] [100 200 300 400 250]

C) [200 200 300 400 250] [100 200 300 400 250]

D) None of these

Solution: (A)

Option A is correct

1. You put bookmark as time.sleep() so that you would know how much the code has “slept” literally
2. You put bookmark as time.time() and check how much time elapses in each code line
3. You put bookmark as datetime.timedelta(), so that you would find out differences of execution times
4. You copy whole code in an Ipython / Jupyter notebook, with each code line as a separate block and write magic function %%timeit in each block

A) 1 & 2

B) 1,2 & 3

C) 1,2 & 4

D) All of the above 

Solution: (C)

Option C is correct

Note: pandas library has been imported as pd In the given file (email.csv), the first three records are empty.

,,,

,,,

,,,

Email_Address,Nickname,Group_Status,Join_Year

[email protected],aa,Owner,2014

[email protected],bb,Member,2015

[email protected],cc,Member,2017

[email protected],dd,Member,2016

A) read_csv(‘email.csv’, skip_rows=3)

B) read_csv(‘email.csv’, skiprows=3)

C) read_csv(‘email.csv’, skip=3)

D) None of these

Solution: (B)

Option B is correct

Now, you want to know whether BMI and Gender would influence the sales.

For this, you want to plot a bar graph as shown below:

The code for this is:

var = df.groupby(['BMI','Gender']).Sales.sum()

var.unstack().plot(kind='bar', method,  color=['red','blue'], grid=False)

A) stacked=True

B) stacked=False

C) stack=False

D) None of these

Solution: (A)

It’s a stacked bar chart.

A) [i for i in City_A if i not in City_B]B) [i for i in City_B if i not in City_A]

C) [i for i in City_A if i in City_B]

D) None of these

Solution: (A)

Option A is correct

Traceback (most recent call last):
File "<input>", line 1, in<module>
UnicodeEncodeError: 'ascii' codec can't encode character.

27) Which of the following would likely correct this error?

Note: pandas has been imported as pdA) pd.read_csv(“temp.csv”, compression=’gzip’)

B) pd.read_csv(“temp.csv”, dialect=’str’)

C) pd.read_csv(“temp.csv”, encoding=’utf-8′)

D) None of these

Solution: (C)

Option C is correct, because encoding should be ‘utf-8’

A) tup(2) = 10

B) tup[2] = 10

C) tup{2} = 10

D) None of these

Solution: (D)

A tuple cannot be updated.

A) urllib2.urlopen(www.abcd.org)B) requests.get(www.abcd.org)

C) Both A and B

D) None of these

Solution: (C)

Option C is correct

html_doc = “””
 <!DOCTYPE html>
 <htmllang="en">
 <head>
 <metacharset="utf-8">
 <metaname="viewport" content="width=device-width">
 <title>udacity/deep-learning: Repo for the Deep Learning Nanodegree Foundations program.</title>
 <linkrel="search" type="application/opensearchdescription+xml" href="/opensearch.xml" title="GitHub">
 <linkrel="fluid-icon" href="https://github.com/fluidicon.png" title="GitHub">
 <metaproperty="fb:app_id" content="1401488693436528">  
 <linkrel="assets" href="https://assets-cdn.github.com/">   
 ... 

“””

30) To read the title of the webpage you are using BeautifulSoup. What is the code for this? 

Hint: You have to extract text in title tag

1. from bs4 import BeautifulSoup
   soup =BeautifulSoup(html_doc,’html.parser’)
   print soup.title.name
2. from bs4 import BeautifulSoup
   soup =BeautifulSoup(html_doc,’html.parser’)
   print soup.title.string
3. from bs4 import BeautifulSoup
   soup=BeautifulSoup(html_doc,’html.parser’)
   print soup.title.get_text
4. None of these

Solution: (B)

Option B is correct
Question Context 31

Imagine, you are given a list of items in a DataFrame as below.

D = [‘A’,’B’,’C’,’D’,’E’,’AA’,’AB’]

Now, you want to apply label encoding on this list for importing and transforming, using LabelEncoder.

from sklearn.preprocessing import LabelEncoder

le = LabelEncoder()

1. array([0, 2, 3, 4, 5, 6, 1])
2. array([0, 3, 4, 5, 6, 1, 2])
3. array([0, 2, 3, 4, 5, 1, 6])
4. Any of the above

Solution: (D)

Option D is correct

Assume, you have defined a data frame which has 2 columns.

import numpy as np

df = pd.DataFrame({'Id':[1,2,3,4],'val':[2,5,np.nan,6]})

A) 0   False
1    False
2   False
3    False

B)  0    False
1    False
2    True
3    False

C) 0 True
1    True
2    True
3    True

D) None of these

Solution: (A)

Option A is correct

Note: HDFS file has been loaded by h5py as hf.

A) hf.key()

B) hf.key

C) hf.keys()

D) None of these

Solution: (C)

Option C is correct

reviews = [‘movie is unwatchable no matter how decent the first half is  . ‘, ‘somewhat funny and well  paced action thriller that has jamie foxx as a hapless  fast  talking hoodlum who is chosen by an overly demanding’, ‘morse is okay as the agent who comes up with the ingenious plan to get whoever did it at all cost .’]

Your task is to find sentiments from the review above. For this, you first write a code to find count of individual words in all the sentences.

counts = Counter()

for i in range(len(reviews)):
    for word in reviews[i].split(value):
        counts[word] += 1

34)What value should we split on to get individual words?

1. ‘ ‘
2. ‘,’
3. ‘.’
4. None of these

Solution: (A)

Option A is correct

import matplotlib.pyplot as plt
plt.plot([1,2,3,4])
plt.show()

1. In line two, write plt.plot([1,2,3,4], width=3)
2. In line two, write plt.plot([1,2,3,4], line_width=3
3. In line two, write plt.plot([1,2,3,4], lw=3)
4. None of these

Solution: (C)

Option C is correct

Note: df is a pandas dataframe

A) df.reset_index(new_index,)

B) df.reindex(new_index,)

C) df.reindex_like(new_index,)

D) None of these

Solution: (A)

Option A is correct

1. crosstab(df_train[‘Pclass’], df_train[‘Survived’])
2. proportion(df_train[‘Pclass’], df_train[‘Survived’])
3. crosstab(df_train[‘Survived’], df_train[‘Pclass’])
4. None of these

Solution: (A)

Option A is correct

1. def generate_ngrams(text, n):
   words = text.split(‘\n’)
   output = [] for i in range(len(words)-n+1):
   append(words[i+1:i+n])
   return output
2. def generate_ngrams(text, n):
   words = text.split()
   output = [] for i in range(len(words)-n+1):
   append(words[i:i+n])
   return output
3. def generate_ngrams(text, n):
   words = text.split()
   output = [] for i in range(len(words)-n+1):
   append(words[i+1:i+n])
   return output
4. None of these

Solution: (B)

Option B is correct

1. df_1.to_csv(‘../data/file.csv’,encoding=’utf-8′,index=True,header=False)
2. df_1.to_csv(‘../data/file.csv’,encoding=’utf-8′,index=False,header=True)
3. df_1.to_csv(‘../data/file.csv’,encoding=’utf-8′,index=False,header=False)
4. None of these

Solution: (C)

Option C is correct

Note: numpy library has been imported as np.

1. def MSE(real_target, predicted_target):
   return np.mean((np.square(real_target) – np.square(predicted_target)))
2. def MSE(real_target, predicted_target):
   return np.mean((real_target – predicted_target)**2)
3. def MSE(real_target, predicted_target):
   return np.sqrt(np.mean((np.square(real_target) – np.square(predicted_target))))
4. None of the above

Solution: (B)

Option B is correct