ValueError when Applying a Function to a Dataframe Python-CodePudding

I have a pandas data frame that consists of mobile numbers.

I want to create a new column to classify each number based on its pattern using regex.

numbers = [539249751,530246444,539246655,539209759,538849098]
  
# Create the pandas DataFrame with column name is provided explicitly
vanity_class= pd.DataFrame(numbers, columns=['MNM_MOBILE_NUMBER'])

I have written a function that iterates through the column MNM_MOBILE_NUMBER. Identifies the pattern of each number using regex. Then, creates a new column MNC_New_Class with the relevant classification.

def vanity_def(MNM_MOBILE_NUMBER):
    if vanity_class.MNM_MOBILE_NUMBER.astype(str).str.match(r'^5(\d)\1{7}') | \
            vanity_class.MNM_MOBILE_NUMBER.astype(str).str.match(r'^5(?!(\d)\1)\d(\d)\2{6}$') | \
            vanity_class.MNM_MOBILE_NUMBER.astype(str).str.match(r'.{2}(?!(\d)\1)\d(\d)\2{5}$') | \
            vanity_class.MNM_MOBILE_NUMBER.astype(str).str.match(r'^\d*(\d)(\d)(?:\1\2){3}\d*$') | \
            vanity_class.MNM_MOBILE_NUMBER.astype(str).str.match(r'^5((\d)\2{3})((\d)\4{3})$') | \
            vanity_class.MNM_MOBILE_NUMBER.astype(str).str.match(r'.{3}(1234567$)'):
        vanity_class['MNC_New_Class'] = 'Diamond'
    elif vanity_class.MNM_MOBILE_NUMBER.astype(str).str.match(r'.{3}(?!(\d)\1)\d(\d)\2{4}$') | \
             vanity_class.MNM_MOBILE_NUMBER.astype(str).str.match(r'^(?!(\d)\1)\d((\d)\3{6})(?!\3)\d$') | \
             vanity_class.MNM_MOBILE_NUMBER.astype(str).str.match(r'\d(\d)\1(\d)\2(\d)\3(\d)\4'):     
        vanity_class['MNC_New_Class'] = 'Gold'
    elif vanity_class.MNM_MOBILE_NUMBER.astype(str).str.match(r"^5(?!(\d)\1)\d((\d)\3{5})(?!\3)\d") | \
            vanity_class.MNM_MOBILE_NUMBER.astype(str).str.match(r"\b\d\d(\d)(?!\1)(\d)\2\2(\d)\3\3\b"):
        vanity_class['MNC_New_Class'] = 'Silver'
    elif vanity_class.MNM_MOBILE_NUMBER.astype(str).str.match(r'.{3}(123456$)') | \
            vanity_class.MNM_MOBILE_NUMBER.astype(str).str.match(r'^5.{3}(?!(\d)\1)\d(\d)\2{3}$') | \
            vanity_class.MNM_MOBILE_NUMBER.astype(str).str.match(r'\d\d(?!(\d)\1)\d((\d)\3{4})(?!\3)\d') | \
            vanity_class.MNM_MOBILE_NUMBER.astype(str).str.match(r'\b\d\d(\d)(\d(00))\2') | \
            vanity_class.MNM_MOBILE_NUMBER.astype(str).str.match(r'^5(\d(000))(\d(000))'):
        vanity_class['MNC_New_Class'] = 'Bronze'
    elif vanity_class.MNM_MOBILE_NUMBER.astype(str).str.match(r"\d\d(?!(\d)\1)\d((\d)\3{3})(?!\3)\d") | \
            vanity_class.MNM_MOBILE_NUMBER.astype(str).str.match(r"\d\d\d(\d\d)(?!\1)(\d)\2(\d)\3\b") | \
            vanity_class.MNM_MOBILE_NUMBER.astype(str).str.match(r"^\d*(\d)(\d)(?:\1\2){2}\d*$"):
        vanity_class['MNC_New_Class'] = 'Special'
    elif vanity_class.MNM_MOBILE_NUMBER.astype(str).str.match(r'.{4}(45678$)') | \
            vanity_class.MNM_MOBILE_NUMBER.astype(str).str.match(r'.{5}(?!(\d)\1)\d(\d){3}') | \
            vanity_class.MNM_MOBILE_NUMBER.astype(str).str.match(r'.{5}(1234$)') | \
            vanity_class.MNM_MOBILE_NUMBER.astype(str).str.match(r'(?!.*(\d)\1(\d)\2(\d)\3).{4}\d(\d)\4(\d)\5') |\
            vanity_class.MNM_MOBILE_NUMBER.astype(str).str.match(r'^\d*(\d)(\d)(?:\1\2){1}\d*$'):
        vanity_class['MNC_New_Class'] = 'Economy'
    else:
        vanity_class['MNC_New_Class'] = 'Non Classified'

Then I wrote a code to apply this function to the dataframe and create a new column.

vanity_class['MNC_New_Class'] = vanity_class['MNM_MOBILE_NUMBER'].apply(vanity_def)

However, I keep getting this error as below

ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

Any advice on how to avoid this error?

Thanks

CodePudding user response：

So quite a few things I see wrong with this. If you look at the documentation for Series.apply, the apply function takes in a single argument and creates a new series where each value is the function applied to the corresponding original value in the original series. So for example, I could have a function like this:

def add_ten_to_each_number(series_value):
    return series_value   10

vanity_class['MNC_New_Class'] = vanity_class['MNM_MOBILE_NUMBER'].apply(add_ten_to_each_number)

This would create a new column with 10 added to each value in the original column. The way it works is apply is called for each value in the series and creates a new series with the returned values from applying the function on each original value .

With that out of the way, your first issue is these kind of lines:

vanity_class['MNC_New_Class'] = 'Diamond'

Instead all you need to do is return 'Diamond' since the apply function will automatically create that new series for you with this value.

The other big issue is that you're not using the passed parameter directly. You're using vanity_class.MNM_MOBILE_NUMBER.astype(str).str.match(). The function parameter itself is already a particular value in vanity_class.MNM_MOBILE_NUMBER since the apply function runs for each value in the series. So you just need to use MNM_MOBILE_NUMBER

The next issue is that you're not converting it to a string correctly because from what I can tell, you think the operation is being performed on the entire series at once. Since this is not the case and it's being performed on each individual value of the series separately (if I'm not wrong), you'll end up with AttributeError: 'int' object has no attribute 'astype'. You can simply do str(MNM_MOBILE_NUMBER) and it should work fine.

This brings me to my final point. Now that you're working with an individual int/str, you can no longer use the .match() function to match regexes as this only works with Series. You'll probably have to use the re python library to handle this.

Hope this helps.

CodePudding user response：

I don't know why you are selecting the each number from the dataframe which I think the error lies there.

To avoid the error and work as intended below is workaround.

def vanity_def(num):
word = None
if (re.match(r'^5(\d)\1{7}',str(num)) or \
        re.match(r'^5(?!(\d)\1)\d(\d)\2{6}$',str(num)) or \
        re.match(r'.{2}(?!(\d)\1)\d(\d)\2{5}$',str(num)) or \
        re.match(r'^\d*(\d)(\d)(?:\1\2){3}\d*$',str(num)) or \
        re.match(r'^5((\d)\2{3})((\d)\4{3})$',str(num)) or \
        re.match(r'.{3}(1234567$)',str(num))):
    word = 'Diamond'
elif (re.match(r'.{3}(?!(\d)\1)\d(\d)\2{4}$',str(num)) or \
         re.match(r'^(?!(\d)\1)\d((\d)\3{6})(?!\3)\d$',str(num)) or \
         re.match(r'\d(\d)\1(\d)\2(\d)\3(\d)\4',str(num))):     
    word = 'Gold'
elif (re.match(r"^5(?!(\d)\1)\d((\d)\3{5})(?!\3)\d",str(num)) or \
        re.match(r"\b\d\d(\d)(?!\1)(\d)\2\2(\d)\3\3\b",str(num))):
    word = 'Silver'
elif (re.match(r'.{3}(123456$)',str(num)) or \
        re.match(r'^5.{3}(?!(\d)\1)\d(\d)\2{3}$',str(num)) or \
        re.match(r'\d\d(?!(\d)\1)\d((\d)\3{4})(?!\3)\d',str(num)) or \
        re.match(r'\b\d\d(\d)(\d(00))\2',str(num)) or \
        re.match(r'^5(\d(000))(\d(000))',str(num))):
    word = 'Bronze'
elif (re.match(r"\d\d(?!(\d)\1)\d((\d)\3{3})(?!\3)\d",str(num)) or \
        re.match(r"\d\d\d(\d\d)(?!\1)(\d)\2(\d)\3\b",str(num)) or \
        re.match(r"^\d*(\d)(\d)(?:\1\2){2}\d*$",str(num))):
    word = 'Special'
elif (re.match(r'.{4}(45678$)',str(num)) or \
        re.match(r'.{5}(?!(\d)\1)\d(\d){3}',str(num)) or \
        re.match(r'.{5}(1234$)',str(num)) or \
        re.match(r'(?!.*(\d)\1(\d)\2(\d)\3).{4}\d(\d)\4(\d)\5',str(num)) or\
        re.match(r'^\d*(\d)(\d)(?:\1\2){1}\d*$',str(num))):
    word = 'Economy'
else:
    word = 'Non Classified'

return word

Output:

    MNM_MOBILE_NUMBER   MNC_New_Class
0   539249751   Economy
1   530246444   Economy
2   539246655   Special
3   539209759   Economy
4   538849098   Economy