update pandas column with another column's values using .loc

Question

I need to conditionally update ColY below if the value for ColX != 0. The difference to other examples is that I need ColY to be replaced by the values from ColX, as opposed to a string

I can replace with a string using .loc when I use the following code:

df1.loc[df1.ColX != 0, 'ColY'] = 'Example'

How can I replace the relevant ColY values with the values from ColX? I have tried things such as the below to no avail

df1.loc[df1.ColX != 0, 'ColY'] = df1.ColX

My original dataframe, df1, is:

ID  ColX   ColY
A   2024   0
B   0      2023
C   2019   0
D   2023   2024

My desired output is:

ID  ColX   ColY
A   2024   2024
B   0      2023
C   2019   2019
D   2023   2023

Answer 1

Just for your convenience, here's another cleaner method in my opinion, using np.where and .ne :

df['ColY'] = np.where(df['ColX'].ne(0), df['ColX'], df['ColY'])

print(df)
  ID  ColX  ColY
0  A  2024  2024
1  B     0  2023
2  C  2019  2019
3  D  2023  2023

Answer 2

The problem with df1.loc[df1.ColX != 0, 'ColY'] = df1.ColX is that you're trying to replace a subset of df1.ColY (ie where df1.ColX != 0 ) with the entire df1.ColX , which has more values.

To copy the correct values conditionally, you also have to apply the same filter to df1.ColX :

df1 = pd.DataFrame(data=[[2024, 0], [0, 2023], [2019, 0], [2023, 2023],], columns=['ColX', 'ColY'])

relevant_cols = (df1.ColX != 0)
df1.loc[relevant_cols, 'ColY'] = df1.loc[relevant_cols, 'ColX']
df1
#   ColX  ColY
# 0  2024  2024
# 1     0  2023
# 2  2019  2019
# 3  2023  2023