Pfthb

I have two dataframes :

df1:

A B C
1 ss 123
2 sv 234
3 sc 333

df2:

A dd xc
1 ss 123

df2 will always have a single row. How to check whether there is a match for that row of df2, in df1?

Using Numpy comparisons with np.all with parameter axis=1 for rows:

df1 = pd.DataFrame('A': [1, 2, 3], 'B': ['ss', 'sv', 'sc'], 'C': [123, 234, 333])
df2 = pd.DataFrame('A': [1], 'dd': ['ss'], 'xc': [123])

df3 = df1.loc[np.all(df1.values == df2.values, axis=1),:]

Or:

df3 = df1.loc[np.all(df1[['B','C']].values == df2[['dd','xc']].values, axis=1),:]

print(df3)
 A B C
0 1 ss 123

Additional to Sandeep's answer, can do:

df1[np.all(df1.values == df2.values,1)].any().any()

For getting a boolean.

Or another way:

df1[(df2.values==df1.values).all(1)].any().any()

Or:

pd.merge(df1,df2).equals(df1)

Note: both output True

Check specific column (same as Sandeep's):

df1[col].isin(df2[col]).any()

How to check whether there is a match for that row of df2, in df1?

You can align columns and then check equality of df1 with the only row of df2:

df2.columns = df1.columns

res = (df1 == df2.iloc[0]).all(1).any() # True

The benefit of this solution is you aren't subsetting df1 (expensive), but instead constructing a Boolean dataframe / array (cheap) and checking if all values in at least one row are True.

This is still not particularly efficient as you are considering every row in df1 rather than stopping when a condition is satisfied. With numeric data, in particular, there are more efficient solutions.