get list of occurrences using pandas

user_2479947 0
Viewed 160

Given a list of animals, like:

animals = ['cat', 'dog', 'hamster', 'dolphin']

and pandas dataframe, df:

id    animals
1     dog,cat
2     dog
3     cat,dolphin
4     cat,dog
5     hamster,dolphin

I want to get a new dataframe showing occurrence of each animal, something like:

animal    ids
cat       1,3,4
dog       1,2,4
hamster   5        
dolphin   3,5

I know I can run a loop and prepare it, but I have the list of over 80,000 words with dataframe of over 1 million rows, so it would take long to do it using loop. Is there an easier and faster method to get the result using dataframe?

pandas dataframe numpy python performance
3 Answers

Let us try get_dummies then dot

df.animals.str.get_dummies(',').T.dot(df.id.astype(str)+',').str[:-1]
Out[307]: 
cat        1,3,4
dog        1,2,4
dolphin      3,5
hamster        5
dtype: object

If would considered the list add reindex

df.animals.str.get_dummies(',').T.dot(df.id.astype(str)+',').str[:-1].reindex(animals)
Out[308]: 
cat        1,3,4
dog        1,2,4
hamster        5
dolphin      3,5
dtype: object
user_79645270

NumPy based one for perf. -

def list_occ(df):
    id_col='id'
    item_col='animals'
    
    sidx = np.argsort(animals)
    s = [i.split(',') for i in df[item_col]]
    d = np.concatenate(s)
    
    p = sidx[np.searchsorted(animals, d, sorter=sidx)]
    C = np.bincount(p, minlength=len(animals))
    
    l = list(map(len,s))
    r = np.repeat(np.arange(len(l)), l)
    v = df[id_col].values[r[np.lexsort((r,p))]]
    
    out = pd.DataFrame({'ids':np.split(v, C[:-1].cumsum())}, index=animals)
    return out

Sample run -

In [41]: df
Out[41]: 
  id          animals
0  1          dog,cat
1  2              dog
2  3      cat,dolphin
3  4          cat,dog
4  5  hamster,dolphin

In [42]: animals
Out[42]: ['cat', 'dog', 'hamster', 'dolphin']

In [43]: list_occ(df)
Out[43]: 
               ids
cat      [1, 3, 4]
dog      [1, 2, 4]
hamster        [5]
dolphin     [3, 5]

Benchmarking

Using the given sample and simply scale up the number of items.

# Setup
N = 100 # scale factor
s = [i.split(',') for i in df['animals']]
df_big = pd.DataFrame({'animals':[[j+str(ID) for j in i] for i in s for ID in range(1,N+1)]})
df_big['id'] = range(1, len(df_big)+1)
animals = np.unique(np.concatenate(df_big.animals)).tolist()
df_big['animals'] = [','.join(i) for i in df_big.animals]
df = df_big

Timings -

# Using given df & scaling it up by replicating elems with progressive IDs
In [9]: N = 100 # scale factor
   ...: s = [i.split(',') for i in df['animals']]
   ...: df_big = pd.DataFrame({'animals':[[j+str(ID) for j in i] for i in s for ID in range(1,N+1)]})
   ...: df_big['id'] = range(1, len(df_big)+1)
   ...: animals = np.unique(np.concatenate(df_big.animals)).tolist()
   ...: df_big['animals'] = [','.join(i) for i in df_big.animals]
   ...: df = df_big

# @BEN_YO's soln-1
In [10]: %timeit df.animals.str.get_dummies(',').T.dot(df.id.astype(str)+',').str[:-1]
163 ms ± 2.94 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

# @BEN_YO's soln-2
In [11]: %timeit df.animals.str.get_dummies(',').T.dot(df.id.astype(str)+',').str[:-1].reindex(animals)
166 ms ± 4.52 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

# @Andy L.'s soln
%timeit (df.astype(str).assign(animals=df.animals.str.split(',')).explode('animals').groupby('animals').id.agg(','.join).reset_index())
13.4 ms ± 74 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

In [12]: %timeit list_occ(df)
2.81 ms ± 101 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
user_32938810

Use str.split, explode and agg.join

df_final = (df.astype(str).assign(animals=df.animals.str.split(','))
                          .explode('animals').groupby('animals').id.agg(','.join)
                          .reset_index())

Out[155]:
   animals     id
0      cat  1,3,4
1      dog  1,2,4
2  dolphin    3,5
3  hamster      5
user_101892140
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