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NumPy - Coggle Diagram

- - - - The logical statement survey_array > 8 evaluates which survey answers were greater than 8, and assigns them a value of 1. np.mean adds all of the 1s up and divides them by the length of survey_array. The resulting output tells us that 20% of responders purchased more than 8 pounds of produce.
  - - - Here’s how we would use NumPy to calculate the 40th percentile of array d:
        .>>> d = np.array([1, 2, 3, 4, 4, 4, 6, 6, 7, 8, 8])
        .>>> np.percentile(d, 40)
        4.00
  - - - Then to find the interquartile range(IQR), we subtract the value of the 25th percentile from the value of the 75th(IQR = Q3 - Q1):
        6.5 - 3.5 = 3
  - - - .>>> nums = np.array([65, 36, 52, 91, 63, 79])
        .>>> np.std(nums)
        17.716909687891082
      - To calculate how many standard deviations are above or below the mean, you must do the following:
        nba_mean = np.mean(nba_data)
        nba_standard_deviation = np.std(nba_data)
        nba_difference = 65 - nba_mean
        num_nba_deviations = nba_difference / nba_standard_deviation
  - - - import numpy as np
        dataset = [3, 5, -2, 49, 10]
        variance = np.var(dataset
  - - - dataset = [50, 10, 4, -3, 4, -20, 2]
        third_quartile = np.quantile(dataset, 0.75)
  - - - For example, if you only wanted the value that split the first 10% of the data apart from the remaining 90%, you could use this code:
        dataset = [5, 10, -20, 42, -9, 10]
        ten_percent = np.quantile(dataset, 0.10)
        However, it would still be useful if you were curious about whether a data point was in the bottom 10% of the dataset.
        
        This result technically isn’t a quantile, because it isn’t splitting the dataset into groups of equal sizes — this value splits the data into one group with 10% of the data and another with 90%.
      - For example, you wanted to get the 5-quantiles you could use this code:
        dataset = [5, 10, -20, 42, -9, 10]
        ten_percent = np.quantile(dataset, [0.2, 0.4, 0.6, 0.8])
- - - - We can transform a regular list into a NumPy array by using np.array() and saving the value to a new variable:
        my_list = [1, 2, 3, 4, 5, 6]
        my_array = np.array(my_list)
  - - - Considered the following CSV(sample.csv):
        34,9,12,11,7
        csv_array = np.genfromtxt('sample.csv', delimiter=',')
        csv_array:
        array([34, 9, 12, 11, 7])
  - - - Taking the square root of each value:
        .>>> np.sqrt(a)
        array([ 1, 1.41421356, 1.73205081, 2, 2.23606798, 2.44948974])
        Squaring each value:
        .>>> a ** 2
        array([ 1, 4, 9, 16, 25, 36])
  - - - .>>> a = np.array([1, 2, 3, 4, 5])
        .>>> b = np.array([6, 7, 8, 9, 10])
        .>>> a + b
        array([ 7, 9, 11, 13, 15])
  - - - We cean store data like this:
        test_1 = np.array([92, 94, 88, 91, 87])
        test_2 = np.array([79, 100, 86, 93, 91])
        test_3 = np.array([87, 85, 72, 90, 92])
        Or in two-dimensional array:
        np.array([[92, 94, 88, 91, 87],
        [ 79, 100, 86, 93, 91],
        [87, 85, 72, 90, 92]])
        
        this is not two-demensional array:
        np.array([[29, 49, 6],
        [77, 1]])
  - - - a = np.array([[32, 15, 6, 9, 14],
        [12, 10, 5, 23, 1],
        [2, 16, 13, 40, 37]])
        select specific elements using their indices:
        .>>> a[2,1]
        16
        select an entire column using ":" as the row index:
        .# selects the first column
        .>>> a[:,0]
        array([32, 12, 2])
  - - - .>>> a = np.array([10, 2, 2, 4, 5, 3, 9, 8, 9, 7])
        .>>> a > 5
        array([True, False, False, False, False, False, True, True, True, True], dtype=bool)
        
        .>>> a[a > 5]
        array([10, 9, 8, 9, 7])
        
        We can also combine logical statements to further specify our criteria. To do so, we place each statement in parentheses and use boolean operators like & (and) and | (or).
        
        .>>> a[(a > 5) | (a < 2)]
        array([10, 9, 8, 9, 7])
- - - - .# Let's generate 10,000 "experiments"
        .# N = 10 shots
        .# P = 0.30 (30% he'll get a free throw)
        a = np.random.binomial(n=10, p=0.30, size=10000)
  - - - Let’s calculate the probability that he makes 4 baskets:
        a = np.random.binomial(10, 0.30, size=10000)
        np.mean(a == 4)
        .>> 0.1973
      - With the large size we chose, the calculated probability should be accurate to about 2 decimal places.*
        So, our basketball player has a roughly 20% chance of making 4 baskets.