Find all columns of dataframe in Pandas whose type

Question

I have a dataframe, df, that has some columns of type float64, while the others are of object. Due to the mixed nature, I cannot use

df.fillna('unknown') #getting error "ValueError: could not convert string to float:"

as the error happened with the columns whose type is float64 (what a misleading error message!)

so I'd wish that I could do something like

for col in df.columns[<dtype == object>]:
    df[col] = df[col].fillna("unknown")

So my question is if there is any such filter expression that I can use with df.columns?

I guess alternatively, less elegantly, I could do:

 for col in df.columns:
        if (df[col].dtype == dtype('O')): # for object type
            df[col] = df[col].fillna('') 
            # still puzzled, only empty string works as replacement, 'unknown' would not work for certain value leading to error of "ValueError: Error parsing datetime string "unknown" at position 0" 

I also would like to know why in the above code replacing '' with 'unknown' the code would work for certain cells but failed with a cell with the error of "ValueError: Error parsing datetime string "unknown" at position 0"

Thanks a lot!

Yu

Answer 1

You can see what the dtype is for all the columns using the dtypes attribute:

In [11]: df = pd.DataFrame([[1, 'a', 2.]])

In [12]: df
Out[12]: 
   0  1  2
0  1  a  2

In [13]: df.dtypes
Out[13]: 
0      int64
1     object
2    float64
dtype: object

In [14]: df.dtypes == object
Out[14]: 
0    False
1     True
2    False
dtype: bool

To access the object columns:

In [15]: df.loc[:, df.dtypes == object]
Out[15]: 
   1
0  a

I think it's most explicit to use (I'm not sure that inplace would work here):

In [16]: df.loc[:, df.dtypes == object] = df.loc[:, df.dtypes == object].fillna('')

Saying that, I recommend you use NaN for missing data.

Answer 2

This is conciser:

# select the float columns
df_num = df.select_dtypes(include=[np.float])
# select non-numeric columns
df_num = df.select_dtypes(exclude=[np.number])