CODEWITHSUNDEEP
pythonpandas
Mohamed Thasin ah
Mohamed Thasin ah

Reputation: 11192

How to randomly insert data for different ranges to different rows

I would like to create a series such that it contains different range of values randomly. Let's say I have a series which contains 12 rows. I would like to pick randomly 4 rows and fill the value randomly between 4 to 10. then again i have to pick another 4 rows and fill the value randomly between -4 to -10. similarly I have to pick rest of all rows and fill the value randomly between 15 to 100. How to achieve this in pandas.

Input:

Col1
0  NaN
1  NaN
2  NaN
3  NaN
4  NaN
5  NaN
6  NaN
7  NaN
8  NaN
9  NaN
10 NaN
11 NaN

Output:

Col1
0  20
1  34
2  -2
3  -7
4  5
5  69
6  -5
7  7
8  97
9  6
10 9
11 -9

So far I tried randomly masking index and filling value randomly using,

df.loc[mask1,'col1']=np.random.randint(4,10, df.shape[0])
df.loc[mask2,'col1']=np.random.randint(-4,-10, df.shape[0])
df.loc[mask3,'col1']=np.random.randint(15,100, df.shape[0])

Is there any other better way to achieve this ?

Upvotes: 1

Views: 37

Answers (2)

Mysak0CZ
Mysak0CZ

Reputation: 964

I think that simplest would be makeing list of indexes and shuffling it.

import random

indexes = list(range(len(data)))  # create list of indexes
random.shuffle(indexes)  # shuffle it

for i in range(len(data)):
    if i < 4:  # first 4 rows
        data[indexes[i]] = random.randint(4, 10)
    elif i < 8:  # another 4 rows
        data[indexes[i]] = random.randint(-4, -10)
    else:  # rest
        data[indexes[i]] = random.randint(15, 100)

EDIT:

import random

indexes = list(range(len(data)))  # create list of indexes
random.shuffle(indexes)  # shuffle it

for i in range(len(data)):
    if i < (len(data)//3):  # first 1/3 rows
        data[indexes[i]] = random.randint(4, 10)
    elif i < (len(data)//3)*2:  # another 1/3 rows
        data[indexes[i]] = random.randint(-10, -4)
    else:  # rest
        data[indexes[i]] = random.randint(15, 100)

I've tested it. What it does is that it fills random 1/3 of elements with first range, 1/2 of left numbers with second range and rest with third range. The indexes are random, because they are picked from "indexes" list with is shuffled. Time complexity is O(n) (linear) where n is lengh of data.

Upvotes: 0

jezrael
jezrael

Reputation: 862641

You can concatenate all values together and then and then use numpy.random.choice:

a = np.r_[np.arange(4,10), np.arange(-4,-10, -1), np.arange(15, 100)]

Or:

a = np.concatenate([np.arange(4,10), np.arange(-4,-10, -1), np.arange(15, 100)])
print (a)
[ 4  5  6  7  8  9 -4 -5 -6 -7 -8 -9 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 65 66 67 68 69 70 71 72 73 74 75 76 77
 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99]

df['col1'] = np.random.choice(a, size=df.shape[0])
print (df)
    col1
0      5
1     65
2     41
3     31
4     86
5      5
6     99
7     42
8     37
9     38
10    -7
11     7

EDIT:

size = int(df.shape[0]/3)
remain = df.shape[0] - 2 * size

a = np.random.randint(4,10, size=size)
b = np.random.randint(-10,-4, size=size)
c = np.random.randint(15,100, size=remain)

d = np.r_[a,b,c]
np.random.shuffle(d)
df['col1'] = d

print (df)
    col1
0      8
1     -7
2     66
3     60
4      8
5     -9
6     24
7     -9
8      7
9      8
10    86
11    -5
12     5
13    -8
14    40

Upvotes: 1

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