I am trying something like this: List append in pandas cell

But the problem is the post is old and everything is deprecated and should not be used anymore.

d = {'col1': ['TEST', 'TEST'], 'col2': [[1, 2], [1, 2]], 'col3': [35, 89]}
df = pd.DataFrame(data=d)

My Dataframe looks like this, were there is the col2 is the one I am interested in. I need to add [0,0] to the lists in col2 for every row in the DataFrame. My real DataFrame is of dynamic shape so I cant just set every cell on its own.

End result should look like this:

I fooled around with df.apply and df.assign but I can't seem to get it to work. I tried:

df['col2']  = [0, 0]

df = df.col2.apply(lambda x: x.append([0,0]))
     Which returns a Series that looks nothing like i need it

df = df.assign(new_column = lambda x: x   list([0, 0))

CodePudding user response：

Not sure if this is the best way to go but, option 2 works with a little modification

import pandas as pd

d = {'col1': ['TEST', 'TEST'], 'col2': [[1, 2], [1, 2]], 'col3': [35, 89]}
df = pd.DataFrame(data=d)
df["col2"] = df["col2"].apply(lambda x: x   [0,0])
print(df)

Firstly, if you want to add all members of an iterable to a list use .extend instead of .append. This doesn't work because the method works inplace and doesn't return anything so "col2" values become None, so use list summation instead. Finally, you want to assign your modified column to the original DataFrame, not override it (this is the reason for the Series return)

CodePudding user response：

One idea is use list comprehension:

df["col2"] = [x   [0,0] for x in  df["col2"]]
   
print (df)
   col1          col2  col3
0  TEST  [1, 2, 0, 0]    35
1  TEST  [1, 2, 0, 0]    89

CodePudding user response：

have you tried the flowing code?,

for val in df['col2']:
val.append(0)

Best Regards, Stan