Create several columns with default values in Salesforce

Question

I have a dataframe that looks like 1000 rows, 10 columns

I want to add 20 columns with only one single value in each column (what I call a default value)

Therefore, my final df would be 1000 rows, with 30 columns

I know that I can do it 30 times by doing:

df['column 11'] = 'default value'
df['column 12'] = 'default value 2'

But I would like to do it in a proper way of coding

I have a dict with my {'column label' : 'defaultvalues'}

How can I do so ?

I've tried pd.insert or pd.concatenate but couldn't find my way through

thanks

regards,

Eric

df.assign(**some_dict_mapping_label_to_value).

jpp
– jpp

2019-04-18 12:57:34 +00:00
Commented Apr 18, 2019 at 12:57 — jpp
– jpp, Commented Apr 18, 2019 at 12:57

Quang Hoang · Accepted Answer · 2019-04-18 12:58:32Z

2

One way to do so:

df_len = len(df)
new_df = pd.DataFrame({col: [val] * df_len for col,val in your_dict.items()})
df = pd.concat((df,new_df), axis=1)

answered Apr 18, 2019 at 12:58

Quang Hoang

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jezrael · Accepted Answer · 2019-04-18 13:03:46Z

1

Generally if possible spaces in keys in dictionary for new columns names use DataFrame constuctor with DataFrame.join:

df = pd.DataFrame({'a':range(5)})
print (df)
   a
0  0
1  1
2  2
3  3
4  4

d = {'A 11' : 's', 'A 12':'c'}
df = df.join(pd.DataFrame(d, index=df.index))
print (df)
   a A 11 A 12
0  0    s    c
1  1    s    c
2  2    s    c
3  3    s    c
4  4    s    c

If no spaces and no numbers in columns names (need valid identifier) is possible use DataFrame.assign:

d = {'A11' : 's', 'A12':'c'}
df = df.assign(**d)
print (df)
   a A11 A12
0  0   s   c
1  1   s   c
2  2   s   c
3  3   s   c
4  4   s   c

Another solution is loop by dictionary and assign:

for k, v in d.items():
    df[k] = v

edited Apr 18, 2019 at 13:03

answered Apr 18, 2019 at 12:57

jezrael

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