set list as value in a column of a pandas dataframe

You'd have to do:

df['new_col'] = [my_list] * len(df)

Example:

In [13]:
df = pd.DataFrame(np.random.randn(5,3), columns=list('abc'))
df

Out[13]:
          a         b         c
0 -0.010414  1.859791  0.184692
1 -0.818050 -0.287306 -1.390080
2 -0.054434  0.106212  1.542137
3 -0.226433  0.390355  0.437592
4 -0.204653 -2.388690  0.106218

In [17]:
df['b'] = [[234]] * len(df)
df

Out[17]:
          a      b         c
0 -0.010414  [234]  0.184692
1 -0.818050  [234] -1.390080
2 -0.054434  [234]  1.542137
3 -0.226433  [234]  0.437592
4 -0.204653  [234]  0.106218

Note that dfs are optimised for scalar values, storing non scalar values defeats the point in my opinion as filtering, looking up, getting and setting become problematic to the point that it becomes a pain

Note that the accepted answer may lead to surprising behavior if you want to modify those lists:

df = pd.DataFrame([1, 2, 3], columns=['a'])
df['lists'] = [[]]* len(df)
df
   a lists
0  1    []
1  2    []
2  3    []
df.loc[df.a == 1, 'lists'][0].append('1')
df
   a lists
0  1   [1]
1  2   [1]
2  3   [1]
# oops

To avoid this you must initialize the lists column with a different list instance per row:

df['lists'] = [[] for r in range(len(df))] # note you can't use a generator
df.loc[df.a == 1, 'lists'][0].append('1')
df
   a lists
0  1   [1]
1  2    []
2  3    []

Don't be fooled by the display there, that 1 is still a string:

df.loc[df.a == 1, 'lists'][0]
['1']

You can use DataFrame.apply:

In [1]:
df = pd.DataFrame([1, 2, 3], columns=['numbers'])
my_list = ['foo', 'bar']
df['lists'] = df.apply(lambda _: my_list, axis=1)
df

Out[1]:
   numbers       lists
0        1  [foo, bar]
1        2  [foo, bar]
2        3  [foo, bar]

Again, be aware that my_list is mutable and shared across the whole dataframe. To avoid that you can make a copy for each row:

df['lists'] = df.apply(lambda _: my_list.copy(), axis=1)