Assign values to multiple columns in Pandas

Question

I have follow simple DataFrame - df :

   0
0  1
1  2
2  3

Once I try to create a new columns and assign some values for them, as example below:

 df['col2', 'col3'] = [(2,3), (2,3), (2,3)] 

I got following structure

   0 (col2, col3)
0  1    (2, 3)
1  2    (2, 3)
2  3    (2, 3)

However, I am looking a way to get as here:

   0 col2, col3
0  1    2,   3
1  2    2,   3
2  3    2,   3

Answer 1

看起来解决方案很简单：

df['col2'], df['col3'] = zip(*[(2,3), (2,3), (2,3)])

Answer 2

There is a convenient solution to joining multiple series to a dataframe via a list of tuples. You can construct a dataframe from your list of tuples before assignment:

df = pd.DataFrame({0: [1, 2, 3]})
df[['col2', 'col3']] = pd.DataFrame([(2,3), (2,3), (2,3)])

print(df)

   0  col2  col3
0  1     2     3
1  2     2     3
2  3     2     3

This is convenient, for example, when you wish to join an arbitrary number of series.

Answer 3

I ran across this issue when trying to apply multiple scalar values to multiple new columns and couldn't find a better way. If I'm missing something blatantly obvious, let me know, but df[['b','c']] = 0 doesn't work. but here's the simplified code:

# Create the "current" dataframe
df = pd.DataFrame({'a':[1,2]})

# List of columns I want to add
col_list = ['b','c']

# Quickly create key : scalar value dictionary
scalar_dict = { c : 0 for c in col_list }

# Create the dataframe for those columns - key here is setting the index = df.index
df[col_list] = pd.DataFrame(scalar_dict, index = df.index)

Or, what appears to be slightly faster is to use .assign() :

df = df.assign(**scalar_dict)

Answer 4

alternatively assign can be used

df.assign(col2 = 2, col3= 3)

https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.assign.html