Reputation: 35646
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.
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
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
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
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
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
Upvotes: 10
Views: 4052
Reputation: 120429
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']
Upvotes: 0
Reputation: 28669
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
Upvotes: 1
Reputation: 35646
There are 2 other approaches to performing this operation:
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.
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]]
Upvotes: 1
Reputation: 35646
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.
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
np.arange
and not df.index
directly?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
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
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
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.
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
Upvotes: 12