1

I've got a column within a df labelled a. It contains a list of int values. I want to return if they are consecutive or not.

I can do it passing in a single list but I want to iterate over each row.

df = pd.DataFrame({'a': [[0,2], [9,11,12], [0,1,2], [10,11,13]]})

def cons(L):
    return all(n-i == L[0] for i,n in enumerate(L))

print(cons(df['a'][0])) # works

df['cons'] = df['a'].apply(cons, axis=1) # error

intended:

              a    cons
0        [0, 2]   False
1   [9, 11, 12]   False 
2     [0, 1, 2]    True
3  [10, 11, 13]   False
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3
  • You're getting an error, TypeError: cons() got an unexpected keyword argument 'axis', because axis isn't a valid keyword argument for Series.apply(), since there's only one axis. In the future, please make a minimal reproducible example including the error, and ask a specific question about it. Stack Overflow is a Q&A site after all. Check out the tour if you haven't already. for more tips, see How to Ask. Commented May 30, 2024 at 13:41
  • Duplicate: Trouble passing in lambda to apply for pandas DataFrame: "TypeError: <lambda>() got an unexpected keyword argument 'axis' ". [I already voted to close as "not reproducible or caused by a typo" before realizing there might be existing questions about this, sorry.] Commented May 30, 2024 at 13:46
  • "a list of string values" - Strings? What you're showing are ints. Maybe you simplified your code but forgot to update the text? Commented May 30, 2024 at 13:51

3 Answers 3

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1

Use list comprehension with test if sorted values are same like generated list by minimal and maximal values of lists:

df['cons'] = [sorted(l) == list(range(min(l), max(l)+1)) for l in df['a']]
#alternative
df['cons'] = df['a'].apply(lambda l: sorted(l) == list(range(min(l), max(l)+1)))

Another idea is use np.diff for test if difference is 1 for all values:

df['cons'] = [np.all(np.diff(sorted(l)) == 1) for l in df['a']]
#alternative
df['cons'] = df['a'].apply(lambda l: np.all(np.diff(sorted(l)) == 1))

If want use your solution:

def cons(L):

    return all(n-i==L[0] for i,n in enumerate(L))

df['cons'] = df['a'].apply(cons)
#alternative
df['cons'] = [all(n-i==L[0] for i,n in enumerate(L)) for L in df['a']]

print (df)
              a   cons
0        [0, 2]  False
1   [9, 11, 12]  False
2     [0, 1, 2]   True
3  [10, 11, 13]  False
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Comments

1

Use itertools.pairwise to compare the successive values:

def cons(L):
    from itertools import pairwise
    return all(a==b-1 for a,b in pairwise(L))

df['cons'] = df['a'].apply(cons)

Output:

              a   cons
0  [10, 11, 12]   True
1  [10, 11, 12]   True
2  [10, 11, 12]   True
3  [10, 11, 13]  False
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3 Comments

sorry. The aim of the question doesn't change but I should have made a more accurate example.
@tonydanza123 I see, this was indeed not obvious ;)
@tonydanza123 I updated the answer with a pythonic solution, you shouldn't need to sort (O(n*logn))
0

You can try that code to solve your issue:

cons=df1.apply(lambda ss:pd.Series(ss[0]).diff().iloc[1:].eq(1).all(),1)
df1.assign(cons=cons)


             a   cons
0        [0, 2]  False
1   [9, 11, 12]  False
2     [0, 1, 2]   True
3  [10, 11, 13]  False
Improve this answer

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