'Random row selection in Pandas dataframe
Is there a way to select random rows from a DataFrame in Pandas.
In R, using the car package, there is a useful function some(x, n) which is similar to head but selects, in this example, 10 rows at random from x.
I have also looked at the slicing documentation and there seems to be nothing equivalent.
Update
Now using version 20. There is a sample method.
df.sample(n)
Solution 1:[1]
With pandas version 0.16.1 and up, there is now a DataFrame.sample method built-in:
import pandas
df = pandas.DataFrame(pandas.np.random.random(100))
# Randomly sample 70% of your dataframe
df_percent = df.sample(frac=0.7)
# Randomly sample 7 elements from your dataframe
df_elements = df.sample(n=7)
For either approach above, you can get the rest of the rows by doing:
df_rest = df.loc[~df.index.isin(df_percent.index)]
Per Pedram's comment, if you would like to get reproducible samples, pass the random_state parameter.
df_percent = df.sample(frac=0.7, random_state=42)
Solution 2:[2]
sample
As of v0.20.0, you can use pd.DataFrame.sample, which can be used to return a random sample of a fixed number rows, or a percentage of rows:
df = df.sample(n=k) # k rows
df = df.sample(frac=k) # int(len(df.index) * k) rows
For reproducibility, you can specify an integer random_state, equivalent to using np.ramdom.seed. So, instead of setting, for example, np.random.seed = 0, you can:
df = df.sample(n=k, random_state=0)
Solution 3:[3]
The best way to do this is with the sample function from the random module,
import numpy as np
import pandas as pd
from random import sample
# given data frame df
# create random index
rindex = np.array(sample(xrange(len(df)), 10))
# get 10 random rows from df
dfr = df.ix[rindex]
Solution 4:[4]
Below line will randomly select n number of rows out of the total existing row numbers from the dataframe df without replacement.
df = df.take(np.random.permutation(len(df))[:n])
Solution 5:[5]
Actually this will give you repeated indices np.random.random_integers(0, len(df), N) where N is a large number.
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
| Solution | Source |
|---|---|
| Solution 1 | |
| Solution 2 | typhon04 |
| Solution 3 | rlmlr |
| Solution 4 | vinzee |
| Solution 5 | rlmlr |
