Create an index with Pandas

Question

suppose I have the following dataframe:

import pandas as pd
df = pd.DataFrame({'return':[10, 28, 32, 50, 25]})

I want to set 25 (last element) to be 100 in a new column and then work backwards doing 50*100/25=200, 32*100/25=32*200/50=128..and so on

   return index
0      25  
1      28   
2      32   128
3      50   200
4      25   100

Is there any way of doing this in -preferably- one line. Thanks!

df['index']=.... ???

df['index'] = df['return'] * 4?

busybear
– busybear

2019-01-11 15:45:32 +00:00
Commented Jan 11, 2019 at 15:45 — busybear
– busybear, Commented Jan 11, 2019 at 15:45

jpp · Accepted Answer · 2019-01-11 15:47:40Z

2

You can use iat to extract the last value of a series:

df['index'] = df['return'] * 100 / df['return'].iat[-1]

print(df)

   return  index
0      10   40.0
1      28  112.0
2      32  128.0
3      50  200.0
4      25  100.0

answered Jan 11, 2019 at 15:47

jpp

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Comments

A l w a y s S u n n y · Accepted Answer · 2019-01-11 16:06:42Z

1

Yeah it's possible with One line. Just use df['return'].iloc[-1] to get the last value of the return column and then simple do like below-

import pandas as pd
df = pd.DataFrame({'return':[10, 28, 32, 50, 25]})
df['index'] = (df['return'] * 100  / df['return'].iloc[-1]).astype(int)
print(df)

Output:

       return  index
0      10     40
1      28    112
2      32    128
3      50    200
4      25    100

answered Jan 11, 2019 at 16:06

A l w a y s S u n n y

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2 Comments

jpp Over a year ago

In the general case you likely don't want to convert to int. Just here we have an int factor because 25 is a divisor of 100.

A l w a y s S u n n y Over a year ago

@jpp agreed sir, I just follow the OP's expected output that's why converted to int

hendriksc1 · Accepted Answer · 2019-01-17 08:26:05Z

0

Maybe like this?

import pandas as pd
df = pd.DataFrame({'return':[10, 28, 32, 50, 25]})

last = df["return"].iloc[-1] 
df['index'] = df['return'].apply(lambda x: x*(100/last))

df

It gives:

    return  index
0   10  40.0
1   28  112.0
2   32  128.0
3   50  200.0
4   25  100.0

edited Jan 17, 2019 at 8:26

answered Jan 11, 2019 at 15:43

hendriksc1

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4 Comments

godfryd Over a year ago

This would be good if instead of hardcoding 25 you would do last = df["return"].iloc[-1] and then df['index'] = df['return'].apply(lambda x: x*(100/last))

Oniropolo Over a year ago

Question: using lambda functions is the same as vectorizing the code? Or does it take as much time as a for loop through all the values?

cs95 Over a year ago

@Oniropolo No, it's the opposite.

Sander van den Oord Over a year ago

If possible don't use apply() with lambda functions. It's slower than vectorized operations.

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