Replace values of a dataframe with the value of another dataframe - python

I have two pandas dataframes
df1 = pd.DataFrame({'A': [1, 3, 5], 'B': [3, 4, 5]})
df2 = pd.DataFrame({'A': [1, 2, 3, 4, 5], 'B': [8, 9, 10, 11, 12], 'C': ['K', 'D', 'E', 'F', 'G']})
The index of both data-frames are 'A'.
How to replace the values of df1's column 'B' with the values of df2 column 'B'?
RESULT of df1:
A B
1 8
3 10
5 12

Maybe dataframe.isin() is what you're searching:
df1['B'] = df2[df2['A'].isin(df1['A'])]['B'].values
print(df1)
Prints:
A B
0 1 8
1 3 10
2 5 12

One of possible solutions:
wrk = df1.set_index('A').B
wrk.update(df2.set_index('A').B)
df1 = wrk.reset_index()
The result is:
A B
0 1 8
1 3 10
2 5 12
Another solution, based on merge:
df1 = df1.merge(df2[['A', 'B']], how='left', on='A', suffixes=['_x', ''])\
.drop(columns=['B_x'])

Related

Find highest two numbers on every row in pandas dataframe and extract the column names

I have a code with multiple columns and I would like to add two more, one for the highest number on the row, and another one for the second highest. However, instead of the number, I would like to show the column name where they are found.
Assume the following data frame:
import pandas as pd
df = pd.DataFrame({'A': [1, 5, 10], 'B': [2, 6, 11], 'C': [3, 7, 12], 'D': [4, 8, 13], 'E': [5, 9, 14]})
To extract the highest number on every row, I can just apply max(axis=1) like this:
df['max1'] = df[['A', 'B', 'C', 'D', 'E']].max(axis = 1)
This gets me the max number, but not the column name itself.
How can this be applied to the second max number as well?
You can sorting values and assign top2 values:
cols = ['A', 'B', 'C', 'D', 'E']
df[['max2','max1']] = np.sort(df[cols].to_numpy(), axis=1)[:, -2:]
print (df)
A B C D E max2 max1
0 1 2 3 4 5 4 5
1 5 6 7 8 9 8 9
2 10 11 12 13 14 13 14
df[['max1','max2']] = np.sort(df[cols].to_numpy(), axis=1)[:, -2:][:, ::-1]
EDIT: For get top2 columns names and top2 values use:
df = pd.DataFrame({'A': [1, 50, 10], 'B': [2, 6, 11],
'C': [3, 7, 12], 'D': [40, 8, 13], 'E': [5, 9, 14]})
cols = ['A', 'B', 'C', 'D', 'E']
#values in numpy array
vals = df[cols].to_numpy()
#columns names in array
cols = np.array(cols)
#get indices that would sort an array in descending order
arr = np.argsort(-vals, axis=1)
#top 2 columns names
df[['top1','top2']] = cols[arr[:, :2]]
#top 2 values
df[['max2','max1']] = vals[np.arange(arr.shape[0])[:, None], arr[:, :2]]
print (df)
A B C D E top1 top2 max2 max1
0 1 2 3 40 5 D E 40 5
1 50 6 7 8 9 A E 50 9
2 10 11 12 13 14 E D 14 13
Another approaches to you can get first max then remove it and get max again to get the second max
import pandas as pd
import numpy as np
df = pd.DataFrame({'A': [1, 15, 10], 'B': [2, 89, 11], 'C': [80, 7, 12], 'D': [4, 8, 13], 'E': [5, 9, 14]})
max1=df.max(axis=1)
maxcolum1=df.idxmax(axis=1)
max2 = df.replace(np.array(df.max(axis=1)),0).max(axis=1)
maxcolum2=df.replace(np.array(df.max(axis=1)),0).idxmax(axis=1)
df2 =pd.DataFrame({ 'max1': max1, 'max2': max2 ,'maxcol1':maxcolum1,'maxcol2':maxcolum2 })
df.join(df2)

How to calculate number of rows between 2 indexes of pandas dataframe

I have the following Pandas dataframe in Python:
import pandas as pd
d = {'col1': [1, 2, 3, 4, 5], 'col2': [6, 7, 8, 9, 10]}
df = pd.DataFrame(data=d)
df.index=['A', 'B', 'C', 'D', 'E']
df
which gives the following output:
col1 col2
A 1 6
B 2 7
C 3 8
D 4 9
E 5 10
I need to write a function (say the name will be getNrRows(fromIndex) ) that will take an index value as input and will return the number of rows between that given index and the last index of the dataframe.
For instance:
nrRows = getNrRows("C")
print(nrRows)
> 2
Because it takes 2 steps (rows) from the index C to the index E.
How can I write such a function in the most elegant way?
The simplest way might be
len(df[row_index:]) - 1
For your information we have built-in function get_indexer_for
len(df)-df.index.get_indexer_for(['C'])-1
Out[179]: array([2], dtype=int64)

Pandas sample by filter criteria

I have a data frame like the one below
d = {'var1': [1, 2, 3, 4], 'var2': [5, 6, 7, 8], 'class': ['a', 'a', 'c', 'b']}
df = pd.DataFrame(data=d)
df
var1 var2 class
0 1 5 a
1 2 6 a
2 3 7 c
3 4 8 b
I would like to be able to change the proportion of the class column. For example I would like to down-sample at random the a class by 50% but keep the number of rows for the other classes the same. the results would be:
df
var1 var2 class
0 1 5 a
1 3 7 c
2 4 8 b
How would this be done.
I used the approach to split the DataFrame into df_selection and df_remaining first.
I then reduced df_selection by REMOVE_PERCENTAGE and merged the resulting DataFrame with df_remaining again.
import numpy as np
import pandas as pd
d = {'var1': [1, 2, 3, 4], 'var2': [5, 6, 7, 8], 'class': ['a', 'a', 'c', 'b']}
df = pd.DataFrame(data=d)
REMOVE_PERCENTAGE = 0.5 # between 0 and 1
df = df.set_index(['class'])
df_selection = df.loc['a'] \
.reset_index()
df_remaining = df.drop('a') \
.reset_index()
rows_to_remove = int(REMOVE_PERCENTAGE * len(df_selection.index))
drop_indices = np.random.choice(df_selection.index, rows_to_remove, replace=False)
df_selection_reduced = df_selection.drop(drop_indices)
df_result = pd.concat([df_selection_reduced, df_remaining]) \
.reset_index(drop=True)
print(df_result)

Convert column suffixes from pandas join into a MultiIndex

I have two pandas DataFrames with (not necessarily) identical index and column names.
>>> df_L = pd.DataFrame({'X': [1, 3],
'Y': [5, 7]})
>>> df_R = pd.DataFrame({'X': [2, 4],
'Y': [6, 8]})
I can join them together and assign suffixes.
>>> df_L.join(df_R, lsuffix='_L', rsuffix='_R')
X_L Y_L X_R Y_R
0 1 5 2 6
1 3 7 4 8
But what I want is to make 'L' and 'R' sub-columns under both 'X' and 'Y'.
The desired DataFrame looks like this:
>>> pd.DataFrame(columns=pd.MultiIndex.from_product([['X', 'Y'], ['L', 'R']]),
data=[[1, 5, 2, 6],
[3, 7, 4, 8]])
X Y
L R L R
0 1 5 2 6
1 3 7 4 8
Is there a way I can combine the two original DataFrames to get this desired DataFrame?
You can use pd.concat with the keys argument, along the first axis:
df = pd.concat([df_L, df_R], keys=['L','R'],axis=1).swaplevel(0,1,axis=1).sort_index(level=0, axis=1)
>>> df
X Y
L R L R
0 1 2 5 6
1 3 4 7 8
For those looking for an answer to the more general problem of joining two data frames with different indices or columns into a multi-index table:
# Prepend a key-level to the column index
# https://stackoverflow.com/questions/14744068
df_L = pd.concat([df_L], keys=["L"], axis=1)
df_R = pd.concat([df_R], keys=["R"], axis=1)
# Join the two dataframes
df = df_L.join(df_R)
# Reorder levels if needed:
df = df.reorder_levels([1,0], axis=1).sort_index(axis=1)
Example:
# Data:
df_L = pd.DataFrame({'X': [1, 3, 5], 'Y': [7, 9, 11]})
df_R = pd.DataFrame({'X': [2, 4], 'Y': [6, 8], 'Z': [10, 12]})
# Result:
# X Y Z
# L R L R R
# 0 1 2.0 7 6.0 10.0
# 1 3 4.0 9 8.0 12.0
# 2 5 NaN 11 NaN NaN
This also solves the special case of the OP with equal indices and columns.
df_L.columns = pd.MultiIndex.from_product([["L", ], df_L.columns])

Replicating rows in pandas dataframe by column value and add a new column with repetition index

My question is similar to one asked here. I have a dataframe and I want to repeat each row of the dataframe k number of times. Along with it, I also want to create a column with values 0 to k-1. So
import pandas as pd
df = pd.DataFrame(data={
'id': ['A', 'B', 'C'],
'n' : [ 1, 2, 3],
'v' : [ 10, 13, 8]
})
what_i_want = pd.DataFrame(data={
'id': ['A', 'B', 'B', 'C', 'C', 'C'],
'n' : [ 1, 2, 2, 3, 3, 3],
'v' : [ 10, 13, 13, 8, 8, 8],
'repeat_id': [0, 0, 1, 0, 1, 2]
})
Command below does half of the job. I am looking for pandas way of adding the repeat_id column.
df.loc[df.index.repeat(df.n)]
Use GroupBy.cumcount and copy for avoid SettingWithCopyWarning:
If you modify values in df1 later you will find that the modifications do not propagate back to the original data (df), and that Pandas does warning.
df1 = df.loc[df.index.repeat(df.n)].copy()
df1['repeat_id'] = df1.groupby(level=0).cumcount()
df1 = df1.reset_index(drop=True)
print (df1)
id n v repeat_id
0 A 1 10 0
1 B 2 13 0
2 B 2 13 1
3 C 3 8 0
4 C 3 8 1
5 C 3 8 2

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