在 Pandas DataFrame 中填充缺失的字段？

Question

There are lots of questions about filling missing values.

I want to fill whole missing columns.

Suppose I have:

df=pd.DataFrame([1,2], columns=['A'])

   A
0  1
1  2

What's the idiomatic way to do something like this?

df.fillmissing(['A','B','C'])

My current code:

for name in colnames:
    if name not in df:
        df[name] = None

This produces:

   A   B   C
0  1 NaN NaN
1  2 NaN NaN

Explanation of output:

In this case A is a no-op, but B and C get added, ie:

• I don't know ahead of time which columns are missing
• I know which columns I want
• I want the most concise code (high performance is not a requirement)

Any suggestions?

Answer 1

Perhaps you need reindex :

df.reindex(['A', 'B', 'C'], axis=1)

   A   B   C
0  1 NaN NaN
1  2 NaN NaN

This fills missing columns with NaN, leaving existing columns as-is.

Answer 2

You could also try transposing it then reindex it:

print(df.T.reindex(['A', 'B', 'C']).T)

Answer 3

Let's say you have the following dataframe df and lookup columns cols :

df=pd.DataFrame([1,2], columns=['A'])
cols = ['A', 'B', 'C']

From there, you can subtract the list of the columns from the dataframe from the list of columns from cols and create the new columns all at once (that don't exist after subtraction) and set to None . Note: You cannot directly subtract lists from each other unless you convert to a set first. Then, enclose [*] around the set to transform to a list:

Method 1: Set Subtraction

df[[*set(cols) - set(df.columns)]] = None
df
Out[1]: 
   A     B     C
0  1  None  None
1  2  None  None

The list comprehension way would be:

Method 2: List Comprehension (similar to your for loop)

df[[col for col in cols if col not in df.columns]] = None
df
Out[1]: 
   A     B     C
0  1  None  None
1  2  None  None