Groupby agg keep blank value - pandas

Let say I have got a dataframe called df
A 10
A 20
15
20
B 10
B 10
The result I want is
A 30
35
B 20


I imagine your blanks are actually NaNs, then use dropna=False:
df.groupby('col1', dropna=False).sum()
If they really are empty strings, then it should work with the default.
Example:
df = pd.DataFrame({'col1': ['A', 'A', float('nan'), float('nan'), 'B', 'B'],
'col2': [10, 20, 15, 20, 10, 10]})
df.groupby('col1', dropna=False).sum()
output:
col2
col1
A 30
B 20
NaN 35


Group by custom group and aggregate columns.
Suppose your dataframe with 2 columns: 'col1' and 'col2':
>>> df
col1 col2
0 A 10 # <- group 1
1 A 20 # <- group 1
2 15 # <- group 2
3 20 # <- group 2
4 B 10 # <- group 3
5 B 10 # <- group 3
grp = df.iloc[:, 0].ne(df.iloc[:, 0].shift()).cumsum()
out = df.groupby(grp, as_index=False).agg({'col1': 'first', 'col2': 'sum'})
Output result:
>>> out
col1 col2
0 A 30
1 35
2 B 20

Related

Convert multiple columns in pandas dataframe to array of arrays

I have the following dataframe:
col1 col2 col3
1 1 2 3
2 4 5 6
3 7 8 9
4 10 11 12
I want to create a new column that will be an array of arrays, that contains a single array consisting of specific columns, casted to float.
So given column names, say "col2" and "col3", the output dataframe would look like this.
col1 col2 col3 new
1 1 2 3 [[2,3]]
2 4 5 6 [[5,6]]
3 7 8 9 [[8,9]]
4 10 11 12 [[11,12]]
What I have so far works, but seems clumsy and I believe there's a better way. I'm fairly new to pandas and numpy.
selected_columns = ["col2", "col3"]
df[selected_columns] = df[selected_columns].astype(float)
df['new'] = df.apply(lambda r: tuple(r[selected_columns]), axis=1)
.apply(np.array)
.apply(lambda r: tuple(r[["new"]]), axis=1)
.apply(np.array)
Appreciate your help, Thanks!

Using agg:
cols = ['col2', 'col3']
df['new'] = df[cols].agg(list, axis=1)
Using numpy:
df['new'] = df[cols].to_numpy().tolist()
Output:
col1 col2 col3 new
1 1 2 3 [2, 3]
2 4 5 6 [5, 6]
3 7 8 9 [8, 9]
4 10 11 12 [11, 12]
2D lists
cols = ['col2', 'col3']
df['new'] = df[cols].agg(lambda x: [list(x)], axis=1)
# or
df['new'] = df[cols].to_numpy()[:,None].tolist()
Output:
col1 col2 col3 new
1 1 2 3 [[2, 3]]
2 4 5 6 [[5, 6]]
3 7 8 9 [[8, 9]]
4 10 11 12 [[11, 12]]

Retrieving values from different columns in Pandas based on a column condition [duplicate]

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']

Get DataFrame values from variable column

I would like to get series with values from column which name is variable and stored in 'COL'.
import pandas as pd
df = pd.DataFrame({'A': [10, 20, 30, 40], 'B': [100, 200, 300, 400], 'COL': ['A', 'B', 'A', 'B']})
df
A B COL
0 10 100 A
1 20 200 B
2 30 300 A
3 40 400 B
What I need as a result:
X
0 10
1 200
2 30
3 400

Use DataFrame.lookup, only necessary all values from COL exist in columns names:
df['X'] = df.lookup(df.index, df['COL'])
print (df)
A B COL X
0 1 10 A 1
1 2 20 B 20
2 3 30 A 3
3 4 40 B 40

this will solve your problem
df["X"]=df.apply(lambda x: x[x["COL"]],axis=1)

Subtract columns from DataFrames with different shapes by looking up based on another column

df2 has more columns and rows than df1. For each row in df2, I want to lookup a corresponding row in df1 based on matching values in one of their columns. From this matching row in df1, I want to subtract a column between df2 and df1. I tried set_index and directly subtracting the dataframes, but that resulted in a lot of NaN.
df1 = pd.DataFrame([[1, 10], [2, 20], [3, 30]],
columns=['A', 'B'])
df2 = pd.DataFrame([[1, 100, 15], [1, 200, 20],
[2, 100, 30], [2, 200, 35],
[3, 100, 50], [3, 200, 55]],
columns=['A', 'X', 'B'])
# For each row in df2, lookup in df1 based on column A, and produce
# difference of values in columnn B.
expected = pd.DataFrame([[1, 100, 5], [1, 200, 10],
[2, 100, 10], [2, 200, 15],
[3, 100, 20], [3, 200, 25]],
columns=['A', 'X', 'B'])
DataFrames:
df1
A B
0 1 10
1 2 20
2 3 30
df2
A X B
0 1 100 15
1 1 200 20
2 2 100 30
3 2 200 35
4 3 100 50
5 3 200 55
expected
A X B
0 1 100 5
1 1 200 10
2 2 100 10
3 2 200 15
4 3 100 20
5 3 200 25

set_index df1 to 'A' and map it back to df2.A. After that do subtraction:
df2['B'] -= df2.A.map(df1.set_index('A').B)
Out[216]:
A X B
0 1 100 5
1 1 200 10
2 2 100 10
3 2 200 15
4 3 100 20
5 3 200 25
Note: In case df2.A has values doesn't exist in df1.A, it will return NaN on that row. I leave it that way because your sample data doesn't specify how to handle it. If you want to keep the value of B the same in that case, you just need to chain .fillna(0) to the end of map or call method subtract with fill_value=0 option
df2['B'] -= df2.A.map(df1.set_index('A').B).fillna(0)

You can use merge also:
df2.merge(df1, on='A').eval('B = B_x - B_y').drop(['B_x','B_y'], axis=1)
Output:
A X B
0 1 100 5
1 1 200 10
2 2 100 10
3 2 200 15
4 3 100 20
5 3 200 25

Dropping Rows with a does not equal condition

I'm attempting to create a new dataframe that drops a certain segment of records from an existing dataframe.
df2=df[df['AgeSeg']!='0-1']
when I look at df2, the records with '0-1' Age Segment are still there.
Output with 0-1 records still in it.
I would expect the new dataframe to not have them. What am I doing wrong?

You can use isin (https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.isin.html)
Simple example:
import pandas as pd
df = pd.DataFrame({'col1': [1, 2, 3, 2, 9], 'col2': [4, 5, 6, 3, 0]})
df = df[df['col1'].isin([2]) != True]
df before:
col1 col2
0 1 4
1 2 5
2 3 6
3 2 3
4 9 0
df after:
col1 col2
0 1 4
2 3 6
4 9 0