Error: float object has no attribute notnull

Question

I have a dataframe:

  a     b     c
0 nan   Y     nan
1  23   N      3
2 nan   N      2
3  44   Y     nan

I wish to have this output:

  a     b     c      d
0 nan   Y     nan   nan
1  23   N      3     96
2 nan   N      2    nan
3  44   Y     nan    44

I wish to have a condition which is when column a is null, then d will be null else if column b is N and column c is not null then column d is equal to column a * column c else column d equal column a

I have done this code but i get the error:

def f4(row):
    if row['a']==np.nan:
       return np.nan
    elif row['b']=="N" & row(row['c'].notnull()):
       return row['a']*row['c']
    else:
       return row['a']

 DF['P1']=DF.apply(f4,axis=1)

can anyone help me point out where is my mistake? I have refer to this and try this but also get the error Creating a new column based on if-elif-else condition

so any way to overcome this, because my row c should be contain nan(null value) — , Jul 03 '17 at 04:15

score 26 · Answer 1 · edited Jun 22 '21 at 10:13

26

Use

pd.isnull(df['Description'][i])

or

pd.isna(df['Description'][i])

edited Jun 22 '21 at 10:13

JiangKui

1,207
1
11
17

answered Dec 17 '18 at 06:57

Max Kleiner

1,442
1
13
14

3

While this code snippet may solve the question, [including an explanation](//s.tk/meta/questions/114762/explaining-entirely-code-based-answers) really helps to improve the quality of your post. Remember that you are answering the question for readers in the future, and those people might not know the reasons for your code suggestion. Please also try not to crowd your code with explanatory comments, this reduces the readability of both the code and the explanations! – Filnor Dec 17 '18 at 07:17
Thanks, right this was too short, heres an add on: For Series and DataFrame, the same type like ndarray of booleans is returned, containing booleans. So pd.notnull(df) and pd.notna(df) results the same. – Max Kleiner Dec 11 '19 at 13:05
Good solution, worked fine when I want to skip rows with NaN in iteration in a for loop: `for i, row in df.iterrows()` , `if pd.notnull(df['Column'][i]):` – Wizhi Jul 13 '20 at 17:05

score 12 · Accepted Answer · answered Jul 03 '17 at 04:16

You don't need apply, use np.where:

df['d'] = np.where(df.a.isnull(),
         np.nan,
         np.where((df.b == "N")&(~df.c.isnull()),
                  df.a*df.c,
                  df.a))

Output:

      a  b    c     d
0   NaN  Y  NaN   NaN
1  23.0  N  3.0  69.0
2   NaN  N  2.0   NaN
3  44.0  Y  NaN  44.0

juanpa.arrivillaga · Answer 3 · 2017-07-03T04:33:24.770

Since you just want Nans to be propagated, multiplying the columns takes care of that for you:

>>> df = pd.read_clipboard()
>>> df
      a  b    c
0   NaN  Y  NaN
1  23.0  N  3.0
2   NaN  N  2.0
3  44.0  Y  NaN
>>> df.a * df.c
0     NaN
1    69.0
2     NaN
3     NaN
dtype: float64
>>>

If you want to do it on a condition, you can use np.where here instead of .apply. all you need is the following:

>>> df
      a  b    c
0   NaN  Y  NaN
1  23.0  N  3.0
2   NaN  N  2.0
3  44.0  Y  NaN
>>> np.where(df.b == 'N', df.a*df.c, df.a)
array([ nan,  69.,  nan,  44.])

This is the default behavior for most operations involving Nan. So, you can simply assign the result of the above:

>>> df['d'] = np.where(df.b == 'N', df.a*df.c, df.a)
>>> df
      a  b    c     d
0   NaN  Y  NaN   NaN
1  23.0  N  3.0  69.0
2   NaN  N  2.0   NaN
3  44.0  Y  NaN  44.0
>>>

Just to elaborate on what this:

np.where(df.b == 'N', df.a*df.c, df.a)

Is doing, you can think of it as "where df.b == 'N', give me the result of df.a * df.c, else, give me just df.a:

>>> np.where(df.b == 'N', df.a*df.c, df.a)
array([ nan,  69.,  nan,  44.])

Also note, if your dataframe were a little different:

>>> df
      a  b    c
0   NaN  Y  NaN
1  23.0  Y  3.0
2   NaN  N  2.0
3  44.0  Y  NaN
>>> df.loc[0,'a'] = 99
>>> df.loc[0, 'b']= 'N'
>>> df
      a  b    c
0  99.0  N  NaN
1  23.0  N  3.0
2   NaN  N  2.0
3  44.0  Y  NaN

Then the following would not be equivalent:

>>> np.where(df.b == 'N', df.a*df.c, df.a)
array([ nan,  69.,  nan,  44.])
>>> np.where((df.b == 'N') & (~df.c.isnull()), df.a*df.c, df.a)
array([ 99.,  69.,  nan,  44.])

So you might want to use the slightly more verbose:

>>> df['d'] = np.where((df.b == 'N') & (~df.c.isnull()), df.a*df.c, df.a)
>>> df
      a  b    c     d
0  99.0  N  NaN  99.0
1  23.0  N  3.0  69.0
2   NaN  N  2.0   NaN
3  44.0  Y  NaN  44.0
>>>

Vaishali · Answer 4 · 2017-07-03T04:23:53.100

You can try

df['d'] = np.where((df.b == 'N') & (pd.notnull(df.c)), df.a*df.c, np.where(pd.notnull(df.a), df.a, np.nan))


    a       b   c      d
0   NaN     Y   NaN    NaN
1   23.0    N   3.0    69.0
2   NaN     N   2.0    NaN
3   44.0    Y   NaN    44.0

See the documentation for pandas notnull, in your current code, you just need to change series.notnull to pd.notnull(series) for it to work. Though np.where should be more efficient

def f4(row):
    if row['a']==np.nan:
        return np.nan
    elif (row['b']=="N") & (pd.notnull(row.c)):
        return row['a']*row['c']
    else:
        return row['a']
df['d']=df.apply(f4,axis=1)

score 1 · Answer 5 · answered Jun 22 '21 at 09:32

1

Using pd.isnull() instead of == np.nan.

Example:

>>> x1 = np.nan
>>> x1 == np.nan
False
>>> pd.isnull(x1)
True
>>> pd.isna(x1)
True

Look this:

The difference between comparison to np.nan and isnull()

answered Jun 22 '21 at 09:32

JiangKui

1,207
1
11
17

Error: float object has no attribute notnull

5 Answers5

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