Update dataframe if certain condition is met

Any help would be greatly appreciated here:

Lets say I have a Pandas DataFrame such as:

Column1 Column2 Column3 Column4  Column5 Column6

0  aaa      abb  jnhs  01/01/2020      40   TEST
1  aba      vvv  jnjh  01/01/2020      34   TEST

I am looking for the best way to be able to test if a certain condition exists, if it does to take
the number on the same row under the heading "column5" below and add 1 to that number. Then to return the full dataframe with just 1 added to the cells in col5 where col6 ="TEST".

The below is the Dataframe result I am looking for:

Column1 Column2 Column3 Column4  Column5 Column6

0  aaa      abb  jnhs  01/01/2020      41   TEST
1  aba      vvv  jnjh  01/01/2020      35   TEST

What I have tried so far:

df['Column5'] = np.where(df['Colum6'] == "TEST", +1, "NA")

This howevever replaces the value in column5 with 1 and does not add 1 to the current value to make them 41 and 35 respectively.

Thank you for any help.

You can look for the rows where the condition is met, then add +1 to Column5

mask = df["Column6"] == "TEST"
df.loc[mask, "Column5"] += 1

This work:

df['Column5'] = np.where(df['Colum6'] == "TEST", df['Column5']+1, "NA")

You can do:

m = df["Column6"].eq("TEST")
df.loc[m, "Column5"] = df.loc[m, "Column5"].add(1)

The pandas documentation recommends using a mask to index the desired rows. And then use the += operator as suggested in another answer or index any other row and just add +1 or any other desired value.

import pandas as pd
import numpy as np

data = {"Index" : [0,1, 2], "A": [1, 1, 1],
        "B": [42, 42, 42], "C": ["test","test", "NA"]}

df = pd.DataFrame(data)

>>>df

Index 	A 	B 	C
0 	0 	1 	42 	test
1 	1 	1 	42 	test
2 	2 	1 	42 	NA

mask = df["C"]=="test"

df.loc[mask, "B"] = df["B"]+1

>>> df

Index 	A 	B 	C
0 	0 	1 	43 	test
1 	1 	1 	43 	test
2 	2 	1 	42 	NA

Try this:

import pandas as pd
result=df["column6"]=="TEST"
df.loc[result,"column5"]=df["column5"]+1