Pandas how can 'replace' work after 'loc'?

Question

I have tried many times, but seems the \'replace\' can NOT work well after use \'loc\'. For example I want to replace the \'conlumn_b\' with an regex for the row that the \'

Answer 1

inplace=True works on the object that it was applied on.

When you call .loc, you're slicing your dataframe object to return a new one.

>>> id(df)
4587248608

And,

>>> id(df.loc[df['conlumn_a'] == 'apple', 'conlumn_b'])
4767716968

Now, calling an in-place replace on this new slice will apply the replace operation, updating the new slice itself, and not the original.

---

Now, note that you're calling replace on a column of int, and nothing is going to happen, because regular expressions work on strings.

Here's what I offer you as a workaround. Don't use regex at all.

m = df['conlumn_a'] == 'apple'
df.loc[m, 'conlumn_b'] = df.loc[m, 'conlumn_b'].replace(11, 'XXX')

df

  conlumn_a conlumn_b
0     apple       123
1    banana        11
2     apple       XXX
3    orange        33

Or, if you need regex based substitution, then -

df.loc[m, 'conlumn_b'] = df.loc[m, 'conlumn_b']\
           .astype(str).replace('^11$', 'XXX', regex=True)

Although, this converts your column to an object column.