Reputation: 9338
A short data frame and I want to create new rows from the existing rows.
What it does now is, each row, each column multiple a random number between 3 to 5:
import pandas as pd
import random
data = {'Price': [59,98,79],
'Stock': [53,60,60],
'Delivery': [11,7,6]}
df = pd.DataFrame(data)
for row in range(df.shape[0]):
new_row = round(df.loc[row] * random.randint(3,5))
new_row.name = 'new row'
df = df.append([new_row])
print (df)
Price Stock Delivery
0 59 53 11
1 98 60 7
2 79 60 6
new row 295 265 55
new row 294 180 21
new row 316 240 24
Is it possible that it can multiple different random numbers to each row? For example:
the 1st row 3 cells multiple (random) [3,4,5]
the 2nd row 3 cells multiple (random) [4,4,3] etc?
Upvotes: 2
Views: 140
Reputation: 4648
You may also generate the multiplication coefficients with the same shape of df
independently, and then concat the element-wise multiplied df * mul
with the original df
:
N.B. This method avoids the notoriously slow .append()
. Benchmark: 10,000 rows finished almost instantly with this method, while .append()
took 40 seconds!
import numpy as np
np.random.seed(111) # reproducibility
mul = np.random.randint(3, 6, df.shape) # 6 not inclusive
df_new = pd.concat([df, df * mul], axis=0).reset_index(drop=True)
Output:
print(df_new)
Price Stock Delivery
0 59 53 11
1 98 60 7
2 79 60 6
3 177 159 33
4 294 300 28
5 395 300 30
print(mul) # check the coefficients
array([[3, 3, 3],
[3, 5, 4],
[5, 5, 5]])
Upvotes: 1
Reputation: 150735
Use np.random.randint(3,6, size=3)
. Actually, you can do at once:
df * np.random.randint(3,6, size=df.shape)
Upvotes: 1
Reputation: 323226
Change the random
to numpy
random.choice
in your for loop
np.random.choice(range(3,5),3)
Upvotes: 1