Round each number in a Python pandas data frame by 2 decimals
This works p_table.apply(pd.Series.round)
however it has no decimal places
Documentation says
import pandas as pd
Series.round(decimals=0, out=None)
i tried this p_table.apply(pd.Series.round(2))
but get this error:
unbound method round() must be called with Series instance as first argument (got int instance instead)
How do I round all elements in the data frame to two decimal places?
[EDIT] Figured it out.
import numpy as np
np.round(p_table, decimals=2)
Solution 1:
Since 0.17.0
version you can do .round(n)
df.round(2)
0 1 2 3
0 0.06 0.67 0.77 0.71
1 0.80 0.56 0.97 0.15
2 0.03 0.59 0.11 0.95
3 0.33 0.19 0.46 0.92
df
0 1 2 3
0 0.057116 0.669422 0.767117 0.708115
1 0.796867 0.557761 0.965837 0.147157
2 0.029647 0.593893 0.114066 0.950810
3 0.325707 0.193619 0.457812 0.920403
Solution 2:
import numpy as np
np.round(p_table, decimals=2)
Solution 3:
Below is a sample reproducible possible way of doing it using pandas round function.
# importing pandas as pd
import pandas as pd
# generate sample dataframe
df = pd.DataFrame(np.random.random([5, 4]), columns =["A", "B", "C"])
# use pandas dataframe.round()function to round off all the decimal values to 2 decimal
df.round(2)
# If you want to customize the round off by individual columns
df.round({"A":1, "B":2, "C":3})
Solution 4:
A B C
0 t 8 10.958904
1 w 2 98.630137
To round column C you can use this:
df['c']=df['c'].apply(lambda x:round(x,2))
The output will be:
A B C
0 t 8 10.96
1 w 2 98.63
Solution 5:
that: data.apply(lambda x: np.round(x, decimals=2))
--- timeit.timer for 100x: 0.00356676544494
is same, but slower, as that: np.round(data,decimals=2)
--- timeit.timer for 100x: 0.000921095
for example both gives:
x y z
Input Sequence
1 5.60 0.85 -6.50
2 5.17 0.72 -6.50
3 5.60 0.89 -6.28
4 5.17 0.76 -6.29
for data:
x y z
Input Sequence
1 5.6000 0.8519 -6.5000
2 5.1730 0.7151 -6.5000
3 5.6000 0.8919 -6.2794
4 5.1724 0.7551 -6.2888
5 5.6000 0.9316 -6.0587