Pandas DataFrame | rpow method
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Pandas DataFrame.rpow(~)
method computes the exponential power of a scalar, sequence, Series or DataFrame and the values in the source DataFrame, that is:
other ** DataFrame
Note that this is just the reverse of pow(~)
method, which does:
DataFrame ** other
Unless you use the parameters axis
, level
and fill_value
, rpow(~)
is equivalent to computing the exponential power using the **
operator.
Parameters
1. other
link | scalar
or sequence
or Series
or DataFrame
The resulting DataFrame will be the exponential power of other
and the source DataFrame.
2. axis
link | int
or string
| optional
Whether to broadcast other
for each column or row of the source DataFrame:
Axis | Description |
---|---|
|
|
|
|
This is only relevant if the shape of the source DataFrame and that of other
does not match. By default, axis=1
.
3. level
| int
or string
| optional
The name or the integer index of the level to consider. This is relevant only if your DataFrame is Multi-index.
4. fill_value
link | float
or None
| optional
The value to replace NaN
before computing the exponential power. If the computation involves two NaN
, then its result would still be NaN
. By default, fill_value=None
.
Return Value
A new DataFrame
computed by the exponential power of the source DataFrame and other
.
Examples
Basic usage
Consider the following DataFrames:
df = pd.DataFrame({"A":[3,4], "B":[5,6]})df_other = pd.DataFrame({"A":[1,1], "B":[2,2]})
[df] | [df_other] A B | A B0 3 5 | 0 1 21 4 6 | 1 1 2
Computing the exponential power of df
and df_other
:
df.rpow(df_other)
A B0 1 321 1 64
Here, we're computing the following element-wise exponential power:
1**3 2**51**4 2**6
Broadcasting
Consider the following DataFrame:
df = pd.DataFrame({"A":[3,4], "B":[5,6]})df
A B0 3 51 4 6
Row-wise
By default, axis=1
, which means that other
will be broadcasted for each row in df
:
df.rpow([1,2]) # axis=1
A B0 1 321 1 64
Here, we're computing the following element-wise exponential power:
1**3 2**51**4 2**6
Column-wise
To broadcast other
for each column in df
, set axis=0
like so:
df.rpow([1,2], axis=0)
A Ba 1 1b 16 64
Here, we're computing the following element-wise exponential power:
1**3 1**52**4 2**6
Specifying fill_value
Consider the following DataFrames:
df = pd.DataFrame({"A":[2,np.NaN], "B":[np.NaN,3]})df_other = pd.DataFrame({"A":[4,5], "B":[np.NaN,np.NaN]})
A B | A B0 2.0 NaN | 0 4 NaN1 NaN 3.0 | 1 5 NaN
By default, when we compute the power using rpow(~)
, any operation with NaN
results in NaN
:
df.rpow(df_other)
A B0 16.0 NaN1 NaN NaN
We can fill the NaN
values before we compute the power by using the fill_value
parameter:
df.rpow(df_other, fill_value=1)
A B0 16.0 NaN1 5.0 1.0
Here, notice when the operation is between two NaN
, the result would still be NaN
regardless of fill_value
.