Update specific rows and columns without creating duplictes-CodePudding

I want to overwrite/update some values in some rows. That is the initial data

   Foo Bar  vals
0    0   a     0
1    1   b     1
2    2   c     2
3    3   d     3
4    4   e     4

I want to give Foo 2 and 4 new vals. This is the expected output.

   Foo Bar  vals
0    0   a     0
1    1   b     1
2    2   c    78
3    3   d     3
4    4   e    63

But that code using DataFrame.update() doesn't work that way and does create duplicated lines.

#!/usr/bin/env python3
import pandas as pd

df = pd.DataFrame({'Foo': range(5), 'Bar': list('abcde'), 'vals': range(5)})
print(df)

df.update(
    pd.DataFrame({'Foo': [2, 4], 'vals': [78, 63]})
)

print(df)

The result is

   Foo Bar  vals
0  2.0   a  78.0
1  4.0   b  63.0
2  2.0   c   2.0
3  3.0   d   3.0
4  4.0   e   4.0

CodePudding user response：

You need to do this:

update_from = pd.DataFrame({'Foo': [2, 4], 'vals': [78, 63]})
m = df['Foo'].isin(update_from['Foo'])
df.loc[m, 'vals'] = df.loc[m, 'Foo'].map(update_from.set_index('Foo')['vals'])

output:

   Foo Bar  vals
0    0   a     0
1    1   b     1
2    2   c    78
3    3   d     3
4    4   e    63

CodePudding user response：

df['column name'] = df['column name'].replace(['old value'],'new value')

what I mean is

import pandas as pd

df = pd.DataFrame({'Foo': range(5), 'Bar': list('abcde'), 'vals':range(5)})

print(df)

df['vals'] = df['vals'].replace([2],[78])

df['vals'] = df['vals'].replace([4],[63])

print(df) # which prints your desired output

CodePudding user response：

Just filter your Foo series with isin() and set the values with lists.

NB. this method does not check for order, only membership. Use @SomeDude's method if order matters.

import pandas as pd

df = pd.DataFrame({'Foo': range(5), 'Bar': list('abcde'), 'vals': range(5)})

df.loc[df["Foo"].isin([2, 4]), "vals"] = [78, 63]
print(df)

gives

   Foo Bar  vals
0    0   a     0
1    1   b     1
2    2   c    78
3    3   d     3
4    4   e    63