Since set_value has been deprecated since version 0.21.0, you should now use at. It can insert a list into a cell without raising a ValueError as loc does. I think this is because at always refers to a single value, while loc can refer to values as well as rows and columns.

df = pd.DataFrame(data={'A': [1, 2, 3], 'B': ['x', 'y', 'z']})df.at[1, 'B'] = ['m', 'n']df =    A   B0   1   x1   2   [m, n]2   3   z

You also need to make sure the column you are inserting into has dtype=object. For example

>>> df = pd.DataFrame(data={'A': [1, 2, 3], 'B': [1,2,3]})>>> df.dtypesA    int64B    int64dtype: object>>> df.at[1, 'B'] = [1, 2, 3]ValueError: setting an array element with a sequence>>> df['B'] = df['B'].astype('object')>>> df.at[1, 'B'] = [1, 2, 3]>>> df   A          B0  1          11  2  [1, 2, 3]2  3          3

Pandas >= 0.21

set_value has been deprecated. You can now use DataFrame.at to set by label, and DataFrame.iat to set by integer position.

Setting Cell Values with at/iat

# Setupdf = pd.DataFrame({'A': [12, 23], 'B': [['a', 'b'], ['c', 'd']]})df    A       B0  12  [a, b]1  23  [c, d]df.dtypesA     int64B    objectdtype: object

If you want to set a value in second row of the "B" to some new list, use DataFrane.at:

df.at[1, 'B'] = ['m', 'n']df    A       B0  12  [a, b]1  23  [m, n]

You can also set by integer position using DataFrame.iat

df.iat[1, df.columns.get_loc('B')] = ['m', 'n']df    A       B0  12  [a, b]1  23  [m, n]

What if I get ValueError: setting an array element with a sequence?

I'll try to reproduce this with:

df    A   B0  12 NaN1  23 NaNdf.dtypesA      int64B    float64dtype: object

df.at[1, 'B'] = ['m', 'n']# ValueError: setting an array element with a sequence.

This is because of a your object is of float64 dtype, whereas lists are objects, so there's a mismatch there. What you would have to do in this situation is to convert the column to object first.

df['B'] = df['B'].astype(object)df.dtypesA     int64B    objectdtype: object

Then, it works:

df.at[1, 'B'] = ['m', 'n']df    A       B0  12     NaN1  23  [m, n]

Possible, But Hacky

Even more wacky, I've found you can hack through DataFrame.loc to achieve something similar if you pass nested lists.

df.loc[1, 'B'] = [['m'], ['n'], ['o'], ['p']]df    A             B0  12        [a, b]1  23  [m, n, o, p]

You can read more about why this works here.

df3.set_value(1, 'B', abc) works for any dataframe. Take care of the data type of column 'B'. Eg. a list can not be inserted into a float column, at that case df['B'] = df['B'].astype(object) can help.