How to remove a row from pandas dataframe based on the length of the column values?

Question

In the following pandas.DataFframe:

df = 
    alfa    beta   ceta
    a,b,c   c,d,e  g,e,h
    a,b     d,e,f  g,h,k
    j,k     c,k,l  f,k,n


        

Answer 1

You can do that test to each row in turn using pandas.DataFrame.apply()

print(df[df['alfa'].apply(lambda x: len(x.split(',')) < 3)])

Gives:

  alfa   beta   ceta
1  a,b  d,e,f  g,h,k
2  j,k  c,k,l  f,k,n

Answer 2

How's this?

df = df[df['alpha'].str.split(',', expand=True)[2].isnull()]

Using expand=True creates a new dataframe with one column for each item in the list. If the list has three or more items, then the third column will have a non-null value.

One problem with this approach is that if none of the lists have three or more items, selecting column [2] will cause a KeyError. Based on this, it's safer to use the solution posted by @Stephen Rauch.

Answer 3

There are at-least two ways to subset the given DF:

1) Split on the comma separator and then compute length of the resulting list:

df[df['alfa'].str.split(",").str.len().lt(3)]

2) Count number of commas and add 1 to the result to account for the last character:

df[df['alfa'].str.count(",").add(1).lt(3)] 

Both produce:

Answer 4

This is the numpy version of @NickilMaveli's answer.

mask = np.core.defchararray.count(df.alfa.values.astype(str), ',') <= 1
pd.DataFrame(df.values[mask], df.index[mask], df.columns)

  alfa   beta   ceta
1  a,b  d,e,f  g,h,k
2  j,k  c,k,l  f,k,n

---

naive timing

Answer 5

Here is an option that is the easiest to remember and still embracing the DataFrame which is the "bleeding heart" of Pandas:

1) Create a new column in the dataframe with a value for the length:

df['length'] = df.alfa.str.len()

2) Index using the new column:

df = df[df.length < 3]

Then the comparison to the above timings, which are not really relevant in this case as the data is very small, and usually is less important than how likely you're going to remember how to do something and not having to interrupt your workflow:

step 1:

%timeit df['length'] = df.alfa.str.len()

359 µs ± 6.83 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

step 2:

df = df[df.length < 3]

627 µs ± 76.9 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

The good news is that when the size grows, time does not grow linearly. For example doing the same operation with 30,000 rows of data takes about 3ms (so 10,000x data, 3x speed increase). Pandas DataFrame is like a train, takes energy to get it going (so not great for small things under absolute comparison, but objectively does not matter much does it...as with small data things are fast anyways).