CODEWITHSUNDEEP
pythonpandas
Gingerhaze
Gingerhaze

Reputation: 706

Round each column in df1 based on corresponding column in df2

df1:

Li Be Sc V Cr Mn
20.1564 -0.0011 -0.1921 0.0343 0.5729 0.1121
19.2871 -0.0027 0.0076 0.066 0.5196 0.0981
0.8693 0.0016 0.1997 0.0317 0.0533 0.014

df2:

Li Be Sc V Cr Mn
2.0 0.050 0.3 0.111 0.50 0.40

I need to round the columns in df1 to the same number of decimals places as the corresponding columns in df2. The issue is that each df contains 40+ columns all needing to be rounded to a specific number of decimal places.

I can do this column by column like

df1["Li"]=df1["Li"].round(1)
df1["Be"]=df1["Be"].round(3)
etc

Is there an easier way to round all the columns in df1 based on the number of decimals in df2

desired output:

Li Be Sc V Cr Mn
20.2 -0.001 -0.2 0.034 0.57 0.11
19.3 -0.003 0 0.066 0.52 0.1
0.9 0.002 0.2 0.032 0.05 0.01

Upvotes: 0

Views: 34

Answers (2)

Corralien
Corralien

Reputation: 120479

You can use Decimal from decimal module to get the exponent part and use .round with a mapping dict to convert all columns:

from decimal import Decimal

exponent = lambda x: abs(Decimal(str(x)).as_tuple().exponent)
rounding = df2.T.squeeze().map(exponent)

out = df1.round(rounding)

Output:

>>> out
     Li   Be   Sc      V   Cr   Mn
0  20.2 -0.0 -0.2  0.034  0.6  0.1
1  19.3 -0.0  0.0  0.066  0.5  0.1
2   0.9  0.0  0.2  0.032  0.1  0.0

>>> rounding
Li    1
Be    2
Sc    1
V     3
Cr    1
Mn    1
Name: 0, dtype: int64

Note: as @mozway suggested you, 0.050 has only 2 decimals not 3 because python doesn't care about trailing zeroes.

Upvotes: 1

Rajkumar Hajgude
Rajkumar Hajgude

Reputation: 167

This could not be a very standard way to convert data to the desired form but it might solve your purpose.

for i in df2.columns:
    a = str(df2[i][0])
    df1[i] = df1[i].round(len(a.split('.')[1]))

Upvotes: 0

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