The operation pandas.DataFrame.lookup is "Deprecated since version 1.2.0", and has since invalidated a lot of previous answers.
This post attempts to function as a canonical resource for looking up corresponding row col pairs in pandas versions 1.2.0 and newer.
Standard LookUp Values With Default Range Index
Given the following DataFrame:
df = pd.DataFrame({'Col': ['B', 'A', 'A', 'B'],
'A': [1, 2, 3, 4],
'B': [5, 6, 7, 8]})
Col A B
0 B 1 5
1 A 2 6
2 A 3 7
3 B 4 8
I would like to be able to lookup the corresponding value in the column specified in Col:
I would like my result to look like:
Col A B Val
0 B 1 5 5
1 A 2 6 2
2 A 3 7 3
3 B 4 8 8
Standard LookUp Values With a Non-Default Index
Non-Contiguous Range Index
Given the following DataFrame:
df = pd.DataFrame({'Col': ['B', 'A', 'A', 'B'],
'A': [1, 2, 3, 4],
'B': [5, 6, 7, 8]},
index=[0, 2, 8, 9])
Col A B
0 B 1 5
2 A 2 6
8 A 3 7
9 B 4 8
I would like to preserve the index but still find the correct corresponding Value:
Col A B Val
0 B 1 5 5
2 A 2 6 2
8 A 3 7 3
9 B 4 8 8
MultiIndex
df = pd.DataFrame({'Col': ['B', 'A', 'A', 'B'],
'A': [1, 2, 3, 4],
'B': [5, 6, 7, 8]},
index=pd.MultiIndex.from_product([['C', 'D'], ['E', 'F']]))
Col A B
C E B 1 5
F A 2 6
D E A 3 7
F B 4 8
I would like to preserve the index but still find the correct corresponding Value:
Col A B Val
C E B 1 5 5
F A 2 6 2
D E A 3 7 3
F B 4 8 8
LookUp with Default For Unmatched/Not-Found Values
Given the following DataFrame
df = pd.DataFrame({'Col': ['B', 'A', 'A', 'C'],
'A': [1, 2, 3, 4],
'B': [5, 6, 7, 8]})
Col A B
0 B 1 5
1 A 2 6
2 A 3 7
3 C 4 8 # Column C does not correspond with any column
I would like to look up the corresponding values if one exists otherwise I'd like to have it default to 0
Col A B Val
0 B 1 5 5
1 A 2 6 2
2 A 3 7 3
3 C 4 8 0 # Default value 0 since C does not correspond
LookUp with Missing Values in the lookup Col
Given the following DataFrame:
Col A B
0 B 1 5
1 A 2 6
2 A 3 7
3 NaN 4 8 # <- Missing Lookup Key
I would like any NaN values in Col to result in a NaN value in Val
Col A B Val
0 B 1 5 5.0
1 A 2 6 2.0
2 A 3 7 3.0
3 NaN 4 8 NaN # NaN to indicate missing
Standard LookUp Values With Any Index
The documentation on Looking up values by index/column labels recommends using NumPy indexing via factorize and reindex as the replacement for the deprecated DataFrame.lookup.
import numpy as np
import pandas as pd
df = pd.DataFrame({'Col': ['B', 'A', 'A', 'B'],
'A': [1, 2, 3, 4],
'B': [5, 6, 7, 8]},
index=[0, 2, 8, 9])
idx, col = pd.factorize(df['Col'])
df['Val'] = df.reindex(columns=col).to_numpy()[np.arange(len(df)), idx]
df
Col A B Val
0 B 1 5 5
1 A 2 6 2
2 A 3 7 3
3 B 4 8 8
factorize is used to convert the column encode the values as an "enumerated type".
idx, col = pd.factorize(df['Col'])
# idx = array([0, 1, 1, 0], dtype=int64)
# col = Index(['B', 'A'], dtype='object')
Notice that B corresponds to 0 and A corresponds to 1. reindex is used to ensure that columns appear in the same order as the enumeration:
df.reindex(columns=col)
B A # B appears First (location 0) A appers second (location 1)
0 5 1
1 6 2
2 7 3
3 8 4
We need to create an appropriate range indexer compatible with NumPy indexing.
The standard approach is to use np.arange based on the length of the DataFrame:
np.arange(len(df))
[0 1 2 3]
Now NumPy indexing will work to select values from the DataFrame:
df['Val'] = df.reindex(columns=col).to_numpy()[np.arange(len(df)), idx]
[5 2 3 8]
*Note: This approach will always work regardless of type of index.
MultiIndex
import numpy as np
import pandas as pd
df = pd.DataFrame({'Col': ['B', 'A', 'A', 'B'],
'A': [1, 2, 3, 4],
'B': [5, 6, 7, 8]},
index=pd.MultiIndex.from_product([['C', 'D'], ['E', 'F']]))
idx, col = pd.factorize(df['Col'])
df['Val'] = df.reindex(columns=col).to_numpy()[np.arange(len(df)), idx]
Col A B Val
C E B 1 5 5
F A 2 6 2
D E A 3 7 3
F B 4 8 8
Why use np.arange and not df.index directly?
Standard Contiguous Range Index
import pandas as pd
df = pd.DataFrame({'Col': ['B', 'A', 'A', 'B'],
'A': [1, 2, 3, 4],
'B': [5, 6, 7, 8]})
idx, col = pd.factorize(df['Col'])
df['Val'] = df.reindex(columns=col).to_numpy()[df.index, idx]
In this case only, there is no error as the result from np.arange is the same as the df.index.
df
Col A B Val
0 B 1 5 5
1 A 2 6 2
2 A 3 7 3
3 B 4 8 8
Non-Contiguous Range Index Error
Raises IndexError:
df = pd.DataFrame({'Col': ['B', 'A', 'A', 'B'],
'A': [1, 2, 3, 4],
'B': [5, 6, 7, 8]},
index=[0, 2, 8, 9])
idx, col = pd.factorize(df['Col'])
df['Val'] = df.reindex(columns=col).to_numpy()[df.index, idx]
df['Val'] = df.reindex(columns=col).to_numpy()[df.index, idx]
IndexError: index 8 is out of bounds for axis 0 with size 4
MultiIndex Error
df = pd.DataFrame({'Col': ['B', 'A', 'A', 'B'],
'A': [1, 2, 3, 4],
'B': [5, 6, 7, 8]},
index=pd.MultiIndex.from_product([['C', 'D'], ['E', 'F']]))
idx, col = pd.factorize(df['Col'])
df['Val'] = df.reindex(columns=col).to_numpy()[df.index, idx]
Raises IndexError:
df['Val'] = df.reindex(columns=col).to_numpy()[df.index, idx]
IndexError: only integers, slices (`:`), ellipsis (`...`), numpy.newaxis (`None`) and integer or boolean arrays are valid indices
LookUp with Default For Unmatched/Not-Found Values
There are a few approaches.
First let's look at what happens by default if there is a non-corresponding value:
import numpy as np
import pandas as pd
df = pd.DataFrame({'Col': ['B', 'A', 'A', 'C'],
'A': [1, 2, 3, 4],
'B': [5, 6, 7, 8]})
# Col A B
# 0 B 1 5
# 1 A 2 6
# 2 A 3 7
# 3 C 4 8
idx, col = pd.factorize(df['Col'])
df['Val'] = df.reindex(columns=col).to_numpy()[np.arange(len(df)), idx]
Col A B Val
0 B 1 5 5.0
1 A 2 6 2.0
2 A 3 7 3.0
3 C 4 8 NaN # NaN Represents the Missing Value in C
If we look at why the NaN values are introduced, we will find that when factorize goes through the column it will enumerate all groups present regardless of whether they correspond to a column or not.
For this reason, when we reindex the DataFrame we will end up with the following result:
idx, col = pd.factorize(df['Col'])
df.reindex(columns=col)
idx = array([0, 1, 1, 2], dtype=int64)
col = Index(['B', 'A', 'C'], dtype='object')
df.reindex(columns=col)
B A C
0 5 1 NaN
1 6 2 NaN
2 7 3 NaN
3 8 4 NaN # Reindex adds the missing column with the Default `NaN`
If we want to specify a default value, we can specify the fill_value argument of reindex which allows us to modify the behaviour as it relates to missing column values:
idx, col = pd.factorize(df['Col'])
df.reindex(columns=col, fill_value=0)
idx = array([0, 1, 1, 2], dtype=int64)
col = Index(['B', 'A', 'C'], dtype='object')
df.reindex(columns=col, fill_value=0)
B A C
0 5 1 0
1 6 2 0
2 7 3 0
3 8 4 0 # Notice reindex adds missing column with specified value `0`
This means that we can do:
idx, col = pd.factorize(df['Col'])
df['Val'] = df.reindex(
columns=col,
fill_value=0 # Default value for Missing column values
).to_numpy()[np.arange(len(df)), idx]
df:
Col A B Val
0 B 1 5 5
1 A 2 6 2
2 A 3 7 3
3 C 4 8 0
*Notice the dtype of the column is int, since NaN was never introduced, and, therefore, the column type was not changed.
LookUp with Missing Values in the lookup Col
factorize has a default na_sentinel=-1, meaning that when NaN values appear in the column being factorized the resulting idx value is -1
import numpy as np
import pandas as pd
df = pd.DataFrame({'Col': ['B', 'A', 'A', np.nan],
'A': [1, 2, 3, 4],
'B': [5, 6, 7, 8]})
# Col A B
# 0 B 1 5
# 1 A 2 6
# 2 A 3 7
# 3 NaN 4 8 # <- Missing Lookup Key
idx, col = pd.factorize(df['Col'])
# idx = array([ 0, 1, 1, -1], dtype=int64)
# col = Index(['B', 'A'], dtype='object')
df['Val'] = df.reindex(columns=col).to_numpy()[np.arange(len(df)), idx]
# Col A B Val
# 0 B 1 5 5
# 1 A 2 6 2
# 2 A 3 7 3
# 3 NaN 4 8 4 <- Value From A
This -1 means that, by default, we'll be pulling from the last column when we reindex. Notice the col still only contains the values B and A. Meaning, that we will end up with the value from A in Val for the last row.
The easiest way to handle this is to fillna Col with some value that cannot be found in the column headers.
Here I use the empty string '':
idx, col = pd.factorize(df['Col'].fillna(''))
# idx = array([0, 1, 1, 2], dtype=int64)
# col = Index(['B', 'A', ''], dtype='object')
Now when I reindex, the '' column will contain NaN values meaning that the lookup produces the desired result:
import numpy as np
import pandas as pd
df = pd.DataFrame({'Col': ['B', 'A', 'A', np.nan],
'A': [1, 2, 3, 4],
'B': [5, 6, 7, 8]})
idx, col = pd.factorize(df['Col'].fillna(''))
df['Val'] = df.reindex(columns=col).to_numpy()[np.arange(len(df)), idx]
df:
Col A B Val
0 B 1 5 5.0
1 A 2 6 2.0
2 A 3 7 3.0
3 NaN 4 8 NaN # Missing as expected
Other Approaches to LookUp
There are 2 other approaches to performing this operation:
apply (Intuitive, but quite slow)
apply can be used on axis=1 in order to use the Column values as the key:
import pandas as pd
df = pd.DataFrame({'Col': ['B', 'A', 'A', 'B'],
'A': [1, 2, 3, 4],
'B': [5, 6, 7, 8]})
df['Val'] = df.apply(lambda row: row[row['Col']], axis=1)
df
Col A B Val
0 B 1 5 5
1 A 2 6 2
2 A 3 7 3
3 B 4 8 8
This operation will work regardless of index type:
import pandas as pd
df = pd.DataFrame({'Col': ['B', 'A', 'A', 'B'],
'A': [1, 2, 3, 4],
'B': [5, 6, 7, 8]},
index=[0, 2, 8, 9])
# Col A B
# 0 B 1 5
# 2 A 2 6
# 8 A 3 7
# 9 B 4 8
df['Val'] = df.apply(lambda row: row[row['Col']], axis=1)
df:
Col A B Val
0 B 1 5 5
2 A 2 6 2
8 A 3 7 3
9 B 4 8 8
When dealing with Missing/Non-Corresponding Values we can use Series.get can be used to remedy this issue:
import numpy as np
import pandas as pd
df = pd.DataFrame({'Col': ['B', 'A', 'C', np.nan],
'A': [1, 2, 3, 4],
'B': [5, 6, 7, 8]})
# Col A B
# 0 B 1 5
# 1 A 2 6
# 2 C 3 7 <- Non Corresponding
# 3 NaN 4 8 <- Missing
df['Val'] = df.apply(lambda row: row.get(row['Col']), axis=1)
Col A B Val
0 B 1 5 5.0
1 A 2 6 2.0
2 C 3 7 NaN # Missing value
3 NaN 4 8 NaN # Missing value
With Default Value
df['Val'] = df.apply(lambda row: row.get(row['Col'], default=-1), axis=1)
Col A B Val
0 B 1 5 5
1 A 2 6 2
2 C 3 7 -1 # Default -1
3 NaN 4 8 -1 # Default -1
apply is extremely flexible and modifications are straightforward, however, the general iterative approach, as well as all the individual Series lookups can become extremely costly in large DataFrames.
get_indexer (limited)
Index.get_indexer can be used to convert the column to index values into an indexer for the DataFrame. This means there is no reason to reindex the DataFrame as the indexer corresponds to the DataFrame as a whole.
import pandas as pd
df = pd.DataFrame({'Col': ['B', 'A', 'A', 'B'],
'A': [1, 2, 3, 4],
'B': [5, 6, 7, 8]})
df['Val'] = df.to_numpy()[df.index, df.columns.get_indexer(df['Col'])]
df
Col A B Val
0 B 1 5 5
1 A 2 6 2
2 A 3 7 3
3 B 4 8 8
This approach is reasonably fast, however, missing values are represented by -1 meaning that if a value is missing it will grab the value from the -1 column (The last column in the DataFrame).
import pandas as pd
df = pd.DataFrame({'A': [1, 2, 3, 4],
'B': [5, 6, 7, 8],
'Col': ['B', 'A', 'A', 'C']})
# A B Col <- Col is now the Last Col
# 0 1 5 B
# 1 2 6 A
# 2 3 7 A
# 3 4 8 C <- Notice Col `C` does not correspond to a Valid Column Header
df['Val'] = df.to_numpy()[df.index, df.columns.get_indexer(df['Col'])]
df:
A B Col Val
0 1 5 B 5
1 2 6 A 2
2 3 7 A 3
3 4 8 C C # <- Value from the last column in the DataFrame (index -1)
It is also notable that not reindexing the DataFrame means converting the entire DataFrame to numpy. This can be very costly if there are many unrelated columns that all need converted:
import numpy as np
import pandas as pd
df = pd.DataFrame({1: 10,
2: 20,
3: 't',
4: 40,
5: np.nan,
'Col': ['B', 'A', 'A', 'B'],
'A': [1, 2, 3, 4],
'B': [5, 6, 7, 8]})
df['Val'] = df.to_numpy()[df.index, df.columns.get_indexer(df['Col'])]
df.to_numpy()
[[10 20 't' 40 nan 'B' 1 5 5]
[10 20 't' 40 nan 'A' 2 6 2]
[10 20 't' 40 nan 'A' 3 7 3]
[10 20 't' 40 nan 'B' 4 8 8]]
Compared to the reindexing approach which only contains columns relevant to the column values:
df.reindex(columns=['B', 'A']).to_numpy()
[[5 1]
[6 2]
[7 3]
[8 4]]
Another option is to build a tuple of the lookup columns, pivot the dataframe, and select the relevant columns with the tuples:
cols = [(ent, ent) for ent in df.Col.unique()]
df.assign(Val = df.pivot(index = None, columns = 'Col')
.reindex(columns = cols)
.ffill(axis=1)
.iloc[:, -1])
Col A B Val
0 B 1 5 5.0
2 A 2 6 2.0
8 A 3 7 3.0
9 B 4 8 8.0
Another possible method is to use melt:
df['value'] = (df.melt('Col', ignore_index=False)
.loc[lambda x: x['Col'] == x['variable'], 'value'])
print(df)
# Output:
Col A B value
0 B 1 5 5
1 A 2 6 2
2 A 3 7 3
3 B 4 8 8
This method also works with Missing/Non-Corresponding Values:
df['value'] = (df.melt('Col', ignore_index=False)
.loc[lambda x: x['Col'] == x['variable'], 'value'])
print(df)
# Output
Col A B value
0 B 1 5 5.0
1 A 2 6 2.0
2 C 3 7 NaN
3 NaN 4 8 NaN
You can replace .loc[...] by query(...) but it's little slower although more expressive:
df['value'] = df.melt('Col', ignore_index=False).query('Col == variable')['value']
Which are the equations used to interpolate a DataFrame in Pandas?
Reading the following link, I couldn't find anything related to them.
https://pandas.pydata.org/docs/reference/api/pandas.DataFrame.interpolate.html
I need exactly this:
But I'm not sure if the interpolate() function is doing the same thing. If that's the case, is there anyway I can change it to work like that?
EDIT: Example of dataframe:
df = pd.DataFrame([[np.nan, 10, np.nan, 20, 17, np.nan, np.nan, 14, np.nan, 10, np.nan],
[5, np.nan, 0, np.nan, np.nan, np.nan, 5, np.nan, 10, np.nan, np.nan],
[3, np.nan, np.nan, np.nan, np.nan, np.nan, 2, np.nan, np.nan, np.nan, np.nan],
[np.nan, np.nan, np.nan, 3, 4, 5, np.nan, 7, 8, 9, np.nan]],
columns=['0', '1', '2', '3', '4', '5', '6', '7', '8', '9', '10'])
Unfortunately, the interpolate method is NOT doing exactly that. However, it is still possible to achieve what you want.
Short answer
df.interpolate(limit=1).mul(~(df.shift(-1).isna() & df.isna())).fillna(0)
Step by step explanation
By default, the interpolate method treats the values as equally spaced. So if you input [0,NaN,10,NaN,NaN,16] for instance, you'll get [0,5,10,12,14,16]. This behavior is controlled by the method parameter of the interpolate function. You don't have to change it in your case.
>>> df = pd.DataFrame([np.nan, 10, np.nan, 20, 17, np.nan, np.nan, 14, np.nan, 10, np.nan], columns=["value"])
>>> df
value
0 NaN
1 10.0
2 NaN
3 20.0
4 17.0
5 NaN
6 NaN
7 14.0
8 NaN
9 10.0
10 NaN
>>> df.interpolate()
value
0 NaN
1 10.0
2 15.0
3 20.0
4 17.0
5 16.0
6 15.0
7 14.0
8 12.0
9 10.0
10 10.0
Now, the default behavior will replace any NaN, but you don't want consecutives NaNs to be replaced, so you need to use the limit parameter.
This parameter limits the number of consecutives NaN that will be replaced, but crucially, if you set the limit to 1, the first NaN of the consecutive NaNs will still be replaced; you don't want that!
>>> df.interpolate(limit=1)
value
0 NaN
1 10.0
2 15.0
3 20.0
4 17.0
5 16.0
6 NaN
7 14.0
8 12.0
9 10.0
10 10.0
To get rid of those first values, you need to know which values are NaN and directly followed by another NaN. Use this :
>>> df.shift(-1).isna() & df.isna()
value
0 False
1 False
2 False
3 False
4 False
5 True
6 False
7 False
8 False
9 False
10 True
You can then multiply your dataframe by the negation (~) of this expression. (Note that n*False = 0 and n*True = n)`
>>> df.interpolate(limit=1).mul(~(df.shift(-1).isna() & df.isna()))
value
0 NaN
1 10.0
2 15.0
3 20.0
4 17.0
5 0.0
6 NaN
7 14.0
8 12.0
9 10.0
10 0.0
Finally, replace the remaining NaN values with 0, using fillna
>>> df.interpolate(limit=1).mul(~(df.shift(-1).isna() & df.isna())).fillna(0)
value
0 0.0
1 10.0
2 15.0
3 20.0
4 17.0
5 0.0
6 0.0
7 14.0
8 12.0
9 10.0
10 0.0
Problem Description
I have a dataframe with a multi-index that is three levels deep (0, 1, 2) and I'd like to join this dataframe with another dataframe which is indexed by level 2 of my original dataframe.
In code, I'd like to turn:
pd.DataFrame(['a', 'b', 'c', 'd']).transpose().set_index([0, 1, 2])
and
pd.DataFrame(['c', 'e']).transpose().set_index(0)
into
pd.DataFrame(['a', 'b', 'c', 'd', 'e']).transpose().set_index([0, 1, 2])
What I've tried
I've tried using swaplevel and then join. Didn't work, though some of the error messages suggested that if only I could set on properly this might work.
I tried concat, but couldn't get this to work either. Not sure it can't work though...
Notes:
I have seen this question in which the answer seems to dodge the question (while solving the problem).
pandas will naturally do this for you if the names of the index levels line up. You can rename the index of the second dataframe and join accordingly.
d1 = pd.DataFrame(['a', 'b', 'c', 'd']).transpose().set_index([0, 1, 2])
d2 = pd.DataFrame(['c', 'e']).transpose().set_index(0)
d1.join(d2.rename_axis(2))
3 1
0 1 2
a b c d e
More Comprehensive Example
d1 = pd.DataFrame([
[1, 2],
[3, 4],
[5, 6],
[7, 8]
], pd.MultiIndex.from_product([['A', 'B'], ['X', 'Y']], names=['One', 'Two']))
d2 = pd.DataFrame([
list('abcdefg')
], ['Y'], columns=list('ABCDEFG'))
d3 = pd.DataFrame([
list('hij')
], ['A'], columns=list('HIJ'))
d1.join(d2.rename_axis('Two')).join(d3.rename_axis('One'))
0 1 A B C D E F G H I J
One Two
A X 1 2 NaN NaN NaN NaN NaN NaN NaN h i j
Y 3 4 a b c d e f g h i j
B X 5 6 NaN NaN NaN NaN NaN NaN NaN NaN NaN NaN
Y 7 8 a b c d e f g NaN NaN NaN