Dropping Rows with a does not equal condition - pandas

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

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]]

flatten dataframe columns to multi-indexed

I want to go from a previously flattened dataframe with single-level columns back to a multi-indexed dataframe.
Here is an example:
import pandas as pd
# Create a sample dataframe with multi-indexed columns
df = pd.DataFrame({('A', 'a'): [1, 2, 3], ('A', 'b'): [4, 5, 6], ('B', 'a'): [7, 8, 9], ('B', 'b'): [10, 11, 12]})
print(df)
The multi-indexed dataframe:
A B
a b a b
0 1 4 7 10
1 2 5 8 11
2 3 6 9 12
Than the flattening:
# Flatten the columns using the to_flat_index() method
df.columns = df.columns.to_flat_index()
print(df)
flattened dataframe with single-level columns:
(A, a) (A, b) (B, a) (B, b)
0 1 4 7 10
1 2 5 8 11
2 3 6 9 12
How to go from a flattened dataframe with single-level columns back to a multi-indexed dataframe?

First Case
Taking the flattened dataframe from the example above:
# Create a new MultiIndex from the columns tuples names
new_columns = pd.MultiIndex.from_tuples(df.columns)
# Assign the new MultiIndex to the columns
df.columns = new_columns
print(df)
Results in:
A B
a b a b
0 1 4 7 10
1 2 5 8 11
2 3 6 9 12
Because the df.columns = df.columns.to_flat_index() method returns a new Index object with a single level, which is the result of flattening the multi-level column names into tuples.
Therefore, to create a new multi-indexed dataframe from a flattened dataframe, you need to extract the tuples of the original multi-indexed column names and pass them to the pd.MultiIndex.from_tuples method.
Second Case
You might encounter differently flattened dataframes like in this example:
import pandas as pd
# Create a sample dataframe with multi-indexed columns
df = pd.DataFrame({('A', 'a'): [1, 2, 3], ('A', 'b'): [4, 5, 6], ('B', 'a'): [7, 8, 9], ('B', 'b'): [10, 11, 12]})
# Flatten the columns
df.columns = ['_'.join(col) for col in df.columns]
print(df)
Which results in:
A_a A_b B_a B_b
0 1 4 7 10
1 2 5 8 11
2 3 6 9 12
In this case you can use the following code to get a multi-indexed dataframe agian:
# Create a new MultiIndex from a list of tuples
new_columns = pd.MultiIndex.from_tuples([tuple(col.split('_')) for col in df.columns])
# Assign the new MultiIndex to the columns
df.columns = new_columns
print(df)
which results in:
A B
a b a b
0 1 4 7 10
1 2 5 8 11
2 3 6 9 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']

convert from one column pandas dataframe to 3 columns based on index

I have:
col1
0 1
1 2
2 3
3 4
4 5
5 6
...
I want, every 3 rows of the original dataframe to become a single row in the new dataframe:
col1 col2 col3
0 1 2 3
1 4 5 6
...
Any suggestions?

The values of the dataframe are an array that can be reshaped using numpy's reshape method. Then, create a new dataframe using the reshaped values. Assuming your existing dataframe is df-
df_2 = pd.DataFrame(df.values.reshape(2, 3), columns=['col1', 'col2', 'col3'])
This will create the new dataframe of two rows and 3 columns.
col1 col2 col3
0 0 1 2
1 3 4 5

You can use set_index and unstack to get the right shape, and add_preffix to change the column name:
print (df.set_index([df.index//3, df.index%3+1])['col1'].unstack().add_prefix('col'))
col1 col2 col3
0 1 2 3
1 4 5 6
in case the original index is not consecutive values but you still want to reshape every 3 rows, replace df.index by np.arange(len(df)) for both in the set_index

you can covert the col in numpy array and then reshape.
In [27]: np.array(df['col1']).reshape( len(df) // 3 , 3 )
Out[27]:
array([[1, 2, 3],
[4, 5, 6]])
In [..] :reshaped_cols = np.array(df['col1']).reshape( len(df) // 3 , 3 )
pd.DataFrame( data = reshaped_cols , columns = ['col1' , 'col2' , 'col3' ] )
Out[30]:
col1 col2 col3
0 1 2 3
1 4 5 6

Multiplying Dataframe by Column Value

I'm currently trying to multiply a dataframe of local currency values and converting it to its relevant Canadian value by multiplying its relevant FX rate.
However, I keep getting this error:
ValueError: operands could not be broadcast together with shapes (12252,) (1021,)
This is the code I'm working with right now. It works when I have a handful rows of data, but keeps getting the ValueError once I use it on the full file (1022 rows of data incl. headers).
import pandas as pd
Local_File = ('RawData.xlsx')
df = pd.read_excel(Local_File, sheet_name = 'Local')
df2 = df.iloc[:,[2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13]].multiply(df['FX Spot Rate'],axis='index')
print (df2)
My dataframe looks something like this with 1022 rows of data (incl. header)
Appreciate any help! Thank you!

df = pd.DataFrame({'A': [1, 2, 3, 3, 1],
'B': [1, 2, 3, 3, 1],
'C': [9, 7, 4, 3, 9]})
A B C
0 1 1 9
1 2 2 7
2 3 3 4
3 3 3 3
4 1 1 9
df.iloc[:,1:] = df.iloc[:,1:].multiply(df['A'][:], axis="index")
df
A B C
0 1 1 9
1 2 4 14
2 3 9 12
3 3 9 9
4 1 1 9