pandas most efficient way to execute arithmetic operations on multiple dataframe columns - python

my first post!
I'm running python 3.8.5 & pandas 1.1.0 on jupyter notebooks.
I want to divide several columns by the corresponding elements in another column of the same dataframe.
For example:
import pandas as pd
df = pd.DataFrame({'a': [2, 3, 4], 'b': [4, 6, 8], 'c':[6, 9, 12]})
df
a b c
0 2 4 6
1 3 6 9
2 4 8 12
I'd like to divide columns 'b' & 'c' by the corresponding values in 'a' and substitute the values in 'b' and 'c' with the result of this division. So the above dataframe becomes:
a b c
0 2 2 3
1 3 2 3
2 4 2 3
I tried
df.iloc[: , 1:] = df.iloc[: , 1:] / df['a']
but this gives:
a b c
0 2 NaN NaN
1 3 NaN NaN
2 4 NaN NaN
I got it working by doing:
for colname in df.columns[1:]:
df[colname] = (df[colname] / df['a'])
Is there a faster way of doing the above by avoiding the for loop?
thanks,
mk

Almost there, use div with axis=0:
df.iloc[:,1:] = df.iloc[:,1:].div(df.a, axis=0)

df.b= df.b/df.a
df.c=df.c/df.a
or
df[['b','c']]=df.apply(lambda x: x[['b','c']]/x.a ,axis=1)

Related

Keep/select columns with the n highest values in last row of a Pandas dataframe

So I have a dataframe as follows:
import pandas as pd
import numpy as np
df = pd.DataFrame(np.array([[1, 2, 3], [4, 3, 6], [7, 2, 9]]),
columns=['a', 'b', 'c'])
df
Output:
a
b
c
1
2
3
4
3
6
7
2
9
I want to select or keep the two columns, with the highest values in the last row. What is the best way to approach?
So in fact I just want to select or keep column 'a' due to value 7 and column 'c' due to value 9.
Try:
df = df[df.iloc[-1].nlargest(2).index]
Output:
c a
0 3 1
1 6 4
2 9 7
If you want to keep original column sequence as well, you can use Index.intersection() together with .nlargest(), as follows:
df[df.columns.intersection(df.iloc[-1].nlargest(2).index, sort=False)]
Result:
a c
0 1 3
1 4 6
2 7 9

Creating new column taking single value from column of another dataframe

I have two dataframes. The first one is df1 = pd.DataFrame({'A': [5, 0], 'B': [2, 4]}) i.e
A B
0 5 2
1 0 4
another one is df2 = pd.DataFrame({'C': [1, 1], 'D': [3, 3]}) i.e
C D
0 1 3
1 1 3
I want want to grab only 4 from df1 and make new column in df2. I have tried this df2['E']=df1['B'][df1['B']==4] and got
C D E
0 1 3 NaN
1 1 3 4.0
I want both rows of df2 to be 4. How can I achieve this? Any help would be immense help.
if the value '4' appears as the last value in your column( like your example), you could do:
df2['E'].fillna(method= 'backfill')
for other methods, have a look here:https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.fillna.html
It is not actually clear what you wanna accomplish here, but I assume you would like to check if there is any "4" in df1 (column B) and then filling all rows in df2 (column E) with "4". Then you could do:
import numpy as np
df2['E'] = np.where(df1['B'].isin([4]).any(), 4, np.nan)
Output:
C D E
0 1 3 4.0
1 1 3 4.0

sum values in different rows and columns dataframe python

My Data Frame
A B C D
2 3 4 5
1 4 5 6
5 6 7 8
How do I add values of different rows and different columns
Column A Row 2 with Column B row 1
Column A Row 3 with Column B row 2
Similarly for all rows
If you only need do this with two columns (and I understand your question well), I think you can use the shift function.
Your data frame (pandas?) is something like:
d = {'A': [2, 1, 5], 'B': [3, 4, 6], 'C': [4, 5, 7], 'D':[5, 6, 8]}
df = pd.DataFrame(data=d)
So, it's possible to create a new data frame with B column shifted:
df2 = df['B'].shift(1)
which gives:
0 NaN
1 3.0
2 4.0
Name: B, dtype: float64
and then, merge this new data with the previous df and, for example, sum the values:
df = df.join(df2, rsuffix='shift')
df['out'] = df['A'] + df['Bshift']
The final output is in out column:
A B C D Bshift out
0 2 3 4 5 NaN NaN
1 1 4 5 6 3.0 4.0
2 5 6 7 8 4.0 9.0
But it's only an intuition, I'm not sure about your question!

Adding a new column to a pandas dataframe with different number of rows

I'm not sure if pandas is made to do this... But I'd like to add a new row to my dataframe with more rows than the existing columns.
Minimal example:
import pandas as pd
df = pd.DataFrame()
df ['a'] = [0,1]
df ['b'] = [0,1,2]
Could someone please explain if this is possible? I'm using a dataframe to store long lists of data and they all have different lengths that I don't necessarily know at the start.
Absolutely possible. Use pd.concat
Demonstration
df1 = pd.DataFrame([[1, 2, 3]])
df2 = pd.DataFrame([[4, 5, 6, 7, 8, 9]])
pd.concat([df1, df2])
df1 looks like
0 1 2
0 1 2 3
df2 looks like
0 1 2 3 4 5
0 4 5 6 7 8 9
pd.concat looks like
0 1 2 3 4 5
0 1 2 3 NaN NaN NaN
0 4 5 6 7.0 8.0 9.0

Pandas: sum DataFrame rows for given columns

I have the following DataFrame:
In [1]:
df = pd.DataFrame({'a': [1, 2, 3],
'b': [2, 3, 4],
'c': ['dd', 'ee', 'ff'],
'd': [5, 9, 1]})
df
Out [1]:
a b c d
0 1 2 dd 5
1 2 3 ee 9
2 3 4 ff 1
I would like to add a column 'e' which is the sum of columns 'a', 'b' and 'd'.
Going across forums, I thought something like this would work:
df['e'] = df[['a', 'b', 'd']].map(sum)
But it didn't.
I would like to know the appropriate operation with the list of columns ['a', 'b', 'd'] and df as inputs.
You can just sum and set param axis=1 to sum the rows, this will ignore none numeric columns:
In [91]:
df = pd.DataFrame({'a': [1,2,3], 'b': [2,3,4], 'c':['dd','ee','ff'], 'd':[5,9,1]})
df['e'] = df.sum(axis=1)
df
Out[91]:
a b c d e
0 1 2 dd 5 8
1 2 3 ee 9 14
2 3 4 ff 1 8
If you want to just sum specific columns then you can create a list of the columns and remove the ones you are not interested in:
In [98]:
col_list= list(df)
col_list.remove('d')
col_list
Out[98]:
['a', 'b', 'c']
In [99]:
df['e'] = df[col_list].sum(axis=1)
df
Out[99]:
a b c d e
0 1 2 dd 5 3
1 2 3 ee 9 5
2 3 4 ff 1 7
If you have just a few columns to sum, you can write:
df['e'] = df['a'] + df['b'] + df['d']
This creates new column e with the values:
a b c d e
0 1 2 dd 5 8
1 2 3 ee 9 14
2 3 4 ff 1 8
For longer lists of columns, EdChum's answer is preferred.
Create a list of column names you want to add up.
df['total']=df.loc[:,list_name].sum(axis=1)
If you want the sum for certain rows, specify the rows using ':'
This is a simpler way using iloc to select which columns to sum:
df['f']=df.iloc[:,0:2].sum(axis=1)
df['g']=df.iloc[:,[0,1]].sum(axis=1)
df['h']=df.iloc[:,[0,3]].sum(axis=1)
Produces:
a b c d e f g h
0 1 2 dd 5 8 3 3 6
1 2 3 ee 9 14 5 5 11
2 3 4 ff 1 8 7 7 4
I can't find a way to combine a range and specific columns that works e.g. something like:
df['i']=df.iloc[:,[[0:2],3]].sum(axis=1)
df['i']=df.iloc[:,[0:2,3]].sum(axis=1)
You can simply pass your dataframe into the following function:
def sum_frame_by_column(frame, new_col_name, list_of_cols_to_sum):
frame[new_col_name] = frame[list_of_cols_to_sum].astype(float).sum(axis=1)
return(frame)
Example:
I have a dataframe (awards_frame) as follows:
...and I want to create a new column that shows the sum of awards for each row:
Usage:
I simply pass my awards_frame into the function, also specifying the name of the new column, and a list of column names that are to be summed:
sum_frame_by_column(awards_frame, 'award_sum', ['award_1','award_2','award_3'])
Result:
Following syntax helped me when I have columns in sequence
awards_frame.values[:,1:4].sum(axis =1)
You can use the function aggragate or agg:
df[['a','b','d']].agg('sum', axis=1)
The advantage of agg is that you can use multiple aggregation functions:
df[['a','b','d']].agg(['sum', 'prod', 'min', 'max'], axis=1)
Output:
sum prod min max
0 8 10 1 5
1 14 54 2 9
2 8 12 1 4
The shortest and simplest way here is to use
df.eval('e = a + b + d')

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