Select and modify a slice in pandas dataframe by integer index

Question

I have a dataframe like the following:

df = pd.DataFrame([[1,2],[10,20],[10,2],[1,40]],columns = ['a','b'])
    a   b
0   1   2
1   10  20
2   10  2
3   1   40

I want to select the b column where a == 1, the following is a classic selecting:

df[df.a == 1].b
    a   b
0   1   2
3   1   40

Then I want to select the ith row of this subdataframe, which isn't the row with index i. There again are several ways, like the following:

df[df.a == 1].b.iloc[[1]]
Output: 
3    40
Name: b, dtype: int64

So far so good. The problem is when I try to modify the value I got there, indeed this selection method yields a copy of the slice of the dataframe, not the object itself. Therefore I can't modify it inplace.

test[test.a == 1].b.iloc[[1]] = 3
SettingWithCopyWarning:
A value is trying to be set on a copy of a slice from a DataFrame

I don't know in which part the 'copy' problem lies, since the two following yield the same problem:

test.iloc[[3]].b = 3
test[test.a == 1].b = 3

So my question is this one: how can I change a value by both a mask selection (conditionally on the a column value) and a row selection (by the rank of the row in the subdataframe, not its index value)?

Answer 1

Use loc with the boolean mask and directly pass the index up:

In[178]:
df.loc[df.loc[df['a'] == 1,'b'].index[1], 'b'] = 3
df

Out[178]: 
    a   b
0   1   2
1  10  20
2  10   2
3   1   3

So here we mask the df using df['a'] == 1, this returns a boolean array and we mask the df and select just column 'b':

In[179]:
df.loc[df['a'] == 1,'b']

Out[179]: 
0    2
3    40
Name: b, dtype: int64

then just subscript the index directly:

In[180]:
df.loc[df['a'] == 1,'b'].index[1]

Out[180]: 3

We can then pass this index label back up to the top-level loc.

This test[test.a == 1].b.iloc[[1]] = 3 is chained indexing which is why the warning is raised.