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6 a. array([1, 1, 1]) |
6 a. array([1, 1, 1]) |
7 #. [1, 1, 1] |
7 #. [1, 1, 1] |
8 #. [1.0, 1.0, 1.0] |
8 #. [1.0, 1.0, 1.0] |
9 #. Error |
9 #. Error |
10 |
10 |
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11 Answer: array([1, 1, 1]) |
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12 |
11 2. What does ones_like([1.2, 3, 4, 5]) produce |
13 2. What does ones_like([1.2, 3, 4, 5]) produce |
12 |
14 |
13 a. [1.2, 3, 4, 5] |
15 a. [1.2, 3, 4, 5] |
14 #. array([1.0, 1.0, 1.0, 1.0]) |
16 #. array([1.0, 1.0, 1.0, 1.0]) |
15 #. array([1, 1, 1, 1]) |
17 #. array([1, 1, 1, 1]) |
16 #. array([1.2, 3, 4, 5]) |
18 #. array([1.2, 3, 4, 5]) |
17 |
19 |
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20 Answer: array([1.0, 1.0, 1.0, 1.0]) |
18 |
21 |
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22 3. The plot of ``u`` vs ``v`` is a bunch of scattered points that show a |
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23 linear trend. How do you find the least square fit line of ``u`` vs ``v``. |
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24 |
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25 Answer:: |
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26 |
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27 A = array(u, ones_like(u)).T |
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28 result = lstsq(A, v) |
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29 m, c = result[0] |
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30 |
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31 lst_line = m * u + c |
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32 |