Find the output for following: [CUET 2022] import numpy as...

Find the output for following: [CUET 2022] import numpy as np pencil np.array( $[10,20,30])$ eraser- np.array $([40,50,70])$ pen= np.array $([11,16,17,18])$ import pandas as pd Stationary pd.DataFrame([pencil, pen, eraser], columns=['AB', 'CD', 'EF",'GH']) print(Stationary) $$ \begin{aligned} &\text { (A) }\\ &\begin{array}{|l|l|l|l|l|} \hline & \text { AB } & \text { CD } & \text { EF } & \text { GH } \\ \hline 0 & 10 & 20 & 30 & \mathrm{NaN} \\ \hline 1 & 11 & 16 & 17 & 18 \\ \hline 2 & 40 & 50 & 70 & \mathrm{NaN} \\ \hline \end{array} \end{aligned} $$ $$ \begin{aligned} &\text { (B) }\\ &\begin{array}{|l|l|l|l|l|} \hline & \mathbf{A B} & \mathbf{C D} & \text { EF } & \text { GH } \\ \hline 0 & 10 & 20 & 30 & \text { No value } \\ \hline 1 & 11 & 16 & 17 & 18 \\ \hline 2 & 40 & 50 & 70 & \text { No value } \\ \hline \end{array} \end{aligned} $$ $$ \begin{aligned} &\text { (C) }\\ &\begin{array}{|l|l|l|l|l|} \hline & \text { AB } & \text { CD } & \text { EF } & \text { GH } \\ \hline 0 & 10 & 20 & 30 & 0 \\ \hline 1 & 11 & 16 & 17 & 18 \\ \hline 2 & 40 & 50 & 70 & 0 \\ \hline \end{array} \end{aligned} $$ $$ \begin{aligned} &\text { (D) }\\ &\begin{array}{|l|l|l|l|l|} \hline & \mathbf{A B} & \mathbf{C D} & \mathbf{E F} & \text { GH } \\ \hline 0 & 10 & 20 & 30 & - \\ \hline 1 & 11 & 16 & 17 & 18 \\ \hline 2 & 40 & 50 & 70 & - \\ \hline \end{array} \end{aligned} $$

Find the output for following:
[CUET 2022]
import numpy as np
pencil np.array( $[10,20,30])$
eraser- np.array $([40,50,70])$
pen= np.array $([11,16,17,18])$
import pandas as pd
Stationary pd.DataFrame([pencil, pen, eraser], columns=['AB', 'CD', 'EF",'GH'])
print(Stationary)
$$
\begin{aligned}
&\text { (A) }\\
&\begin{array}{|l|l|l|l|l|}
\hline & \text { AB } & \text { CD } & \text { EF } & \text { GH } \\
\hline 0 & 10 & 20 & 30 & \mathrm{NaN} \\
\hline 1 & 11 & 16 & 17 & 18 \\
\hline 2 & 40 & 50 & 70 & \mathrm{NaN} \\
\hline
\end{array}
\end{aligned}
$$
$$
\begin{aligned}
&\text { (B) }\\
&\begin{array}{|l|l|l|l|l|}
\hline & \mathbf{A B} & \mathbf{C D} & \text { EF } & \text { GH } \\
\hline 0 & 10 & 20 & 30 & \text { No value } \\
\hline 1 & 11 & 16 & 17 & 18 \\
\hline 2 & 40 & 50 & 70 & \text { No value } \\
\hline
\end{array}
\end{aligned}
$$
$$
\begin{aligned}
&\text { (C) }\\
&\begin{array}{|l|l|l|l|l|}
\hline & \text { AB } & \text { CD } & \text { EF } & \text { GH } \\
\hline 0 & 10 & 20 & 30 & 0 \\
\hline 1 & 11 & 16 & 17 & 18 \\
\hline 2 & 40 & 50 & 70 & 0 \\
\hline
\end{array}
\end{aligned}
$$
$$
\begin{aligned}
&\text { (D) }\\
&\begin{array}{|l|l|l|l|l|}
\hline & \mathbf{A B} & \mathbf{C D} & \mathbf{E F} & \text { GH } \\
\hline 0 & 10 & 20 & 30 & - \\
\hline 1 & 11 & 16 & 17 & 18 \\
\hline 2 & 40 & 50 & 70 & - \\
\hline
\end{array}
\end{aligned}
$$

Key Concepts

Representation of Missing Data (NaN)

NaN, which stands for 'Not a Number', is used in Pandas to represent missing or undefined values. It allows for a uniform way to denote gaps in data, enabling downstream processes such as data cleaning, aggregation, and analysis to manage incomplete datasets effectively.

NumPy Arrays

NumPy arrays are a core component of numerical computing in Python, providing a fast and memory-efficient way to store and manipulate homogeneous data. They are frequently used as inputs to Pandas DataFrames, bridging numerical computations and data analysis, and handling operations that require high-performance array processing.

Pandas DataFrame Construction

A Pandas DataFrame is a two-dimensional, tabular data structure that allows you to organize data in rows and columns with labels. It can be built from various data inputs such as lists, dictionaries, or NumPy arrays, and explicit column names can be provided to assign meaning to the data. This feature is essential for data analysis and manipulation tasks.

Handling Inconsistent Data Lengths

When constructing a DataFrame from multiple sequences or arrays that have different lengths, Pandas fills the missing positions with a placeholder (NaN) to maintain a consistent rectangular structure. This automatic handling ensures that all rows have the same number of columns even when the input data is irregular.

Transcript

00:01 In this question we are provided with three vectors, that is, vector a is minus 2i plus 3j, vector b is 2i minus 3j, and vector c is minus 5j.

00:17 And we have to find certain quantities among which first one of is 2a minus 4b.

00:25 So you can see that we have to make twice of vector a and minus four times of vector b okay so you can see here twice of vector a that is having components minus two and three and minus four times of vector b that is two minus three now we have to make a scalar multiplication here so i should get here then minus of four and six and then we can see here we multiply by four so we get here eight and minus twelve now we have to subtract these vectors so when i'm going to subtract minus 4 and 8 i get here minus 12 and 6 plus 12 that is 18 okay so i can write this vector as minus 12 i and plus of 18 j so this is the answer for the first one now i move on the second one second one is we have to find the dot product of a and b that is dot product of minus 2 .3 with 2.

01:24 Minus 3 okay so i have to multiply the components means minus 2 with 2 and 3 with minus 3 after multiplying it i get minus 13 this is the answer for the second part now i move on the third part in the third part we have to find the value of a dot b plus c so a is minus 2 .3 and dot product with sum of b and c so i'm going to add b and c b c is 0 comma minus 5 okay now first of all we have to add because we have to solve the brackets so after adding these we get here 2 comma minus sorry that is minus 3 okay minus 3 minus 5 that is minus 8 now i have to make the scalar product or dot product so i have to multiply the components minus 2 with 2 and 3 with minus 8 after multiplying them i get minus 28 so this is the answer for third part now i'm going to move on the fourth part in the fourth part we have to find the value of minus 2a plus 3b and with dot product 5c so i have to make minus 2 times of a that is minus 2 .3 and plus 3 times of b b is 2.

02:48 Minus 3 and this has to be made dot product with 5 times of c that is 5 times of 0 comma minus 5 okay now see what we get here so first of all we have to make the addition here after waiting these vectors i should get here 10 comma minus 15 and then dot product with the 5 times of the second vector that is 0.

03:15 Minus 15 okay you can see after making the dot product i get 10 multiplied by 0 and minus 15 multiplied by minus 15 that is equals to we can see here sorry minus 25 was there okay because 5 into 5 turns out into 25 so we should have your 25 and after multiplying them we get 375 so this is the answer for the fourth part now i move on the fifth part in the fifth part it is told that we have to find the magnitude of a and multiplied by c dot a okay so magnitude of a should be square root of minus 2 square plus 3 square and c dot a c is 0.

04:00 Minus 5 and dot product with a a is minus 2 .3 so we can see this magnitude turns out into square root of 13 and we have to make the dot product here when making dot product i get here 0 and minus of 15 and after multiplying them i get minus 15 square root 3 13 this is the answer for the fifth part now i move on the 6th part in the 6th part, we have to find the value of b .b minus modulars of b.

04:32 B .b means 2.

04:34 Minus 3, dot product with 2.

04:36 Minus 3 and minus magnitude of b means square root of 2 square and minus 3 square and minus 3 % of 2 %.

04:42 Now this turns out into 2 into 2 and minus 3 into minus 2 % and minus square root of 13...

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