Set value for particular cell in pandas DataFrame using index

Question

I\'ve created a Pandas DataFrame

df = DataFrame(index=[\'A\',\'B\',\'C\'], columns=[\'x\',\'y\'])

and got this

    x    y
A  NaN         


        

Answer 1

you can use .iloc.

df.iloc[[2], [0]] = 10

Answer 2

In my example i just change it in selected cell

    for index, row in result.iterrows():
        if np.isnan(row['weight']):
            result.at[index, 'weight'] = 0.0

'result' is a dataField with column 'weight'

Answer 3

If you want to change values not for whole row, but only for some columns:

x = pd.DataFrame({'A': [1, 2, 3], 'B': [4, 5, 6]})
x.iloc[1] = dict(A=10, B=-10)

Answer 4

RukTech's answer, df.set_value('C', 'x', 10), is far and away faster than the options I've suggested below. However, it has been slated for deprecation.

Going forward, the recommended method is .iat/.at.

---

Why df.xs('C')['x']=10 does not work:

df.xs('C') by default, returns a new dataframe with a copy of the data, so

df.xs('C')['x']=10

modifies this new dataframe only.

df['x'] returns a view of the df dataframe, so

df['x']['C'] = 10

modifies df itself.

Warning: It is sometimes difficult to predict if an operation returns a copy or a view. For this reason the docs recommend avoiding assignments with "chained indexing".

---

So the recommended alternative is

df.at['C', 'x'] = 10

which does modify df.

---

In [18]: %timeit df.set_value('C', 'x', 10)
100000 loops, best of 3: 2.9 µs per loop

In [20]: %timeit df['x']['C'] = 10
100000 loops, best of 3: 6.31 µs per loop

In [81]: %timeit df.at['C', 'x'] = 10
100000 loops, best of 3: 9.2 µs per loop

Answer 5

Try using df.loc[row_index,col_indexer] = value

Answer 6

.iat/.at is the good solution. Supposing you have this simple data_frame:

   A   B   C
0  1   8   4 
1  3   9   6
2  22 33  52

if we want to modify the value of the cell [0,"A"] u can use one of those solution :

1. df.iat[0,0] = 2
2. df.at[0,'A'] = 2

And here is a complete example how to use iat to get and set a value of cell :

def prepossessing(df):
  for index in range(0,len(df)): 
      df.iat[index,0] = df.iat[index,0] * 2
  return df

y_train before :

    0
0   54
1   15
2   15
3   8
4   31
5   63
6   11

y_train after calling prepossessing function that iat to change to multiply the value of each cell by 2:

     0
0   108
1   30
2   30
3   16
4   62
5   126
6   22