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VisualizaçõesCount of elements in lists within pandas data frame
I need to get the frequency of each element in a list when the list is in a pandas data frame columns
In data:
din=pd.DataFrame({'x':[['a','b','c'],['a','e','d', 'c']]})`
x
0 [a, b, c]
1 [a, e, d, c]
Desired Output:
f x
0 2 a
1 1 b
2 2 c
3 1 d
4 1 e
I can expand the list into rows and then perform a group by but this data could be large ( million plus records ) and was wondering if there is a more efficient/direct way.
Thanks
·
Santiago Trujillo
3 Respostas
Responde à pergunta0
First flatten values of lists and then count by value_counts or size or Counter:
a = pd.Series([item for sublist in din.x for item in sublist])
Or:
a = pd.Series(np.concatenate(din.x))
df = a.value_counts().sort_index().rename_axis('x').reset_index(name='f')
Or:
df = a.groupby(a).size().rename_axis('x').reset_index(name='f')
from collections import Counter
from itertools import chain
df = pd.Series(Counter(chain(*din.x))).sort_index().rename_axis('x').reset_index(name='f')
print (df)
x f
0 a 2
1 b 1
2 c 2
3 d 1
4 e 1
·
Santiago Trujillo
Relatório
0
You can also have an one liner like this:
df = pd.Series(sum([item for item in din.x], [])).value_counts()
·
Santiago Trujillo
Relatório
0
It is actually pretty easy with flattened lists and counters
from matplotlib.cbook import flatten
from collections import Counter
din={'x':[['a','b','c'],['a','e','d', 'c']]}
for a,i in din.items() :
u=pd.DataFrame.from_dict(dict(Counter([*flatten(i)])), orient ='index').reset_index().rename(columns ={'index':a,0:str(a)+'_number'})
output:
However if din has several keys and values you will need a function to do the same trick
from matplotlib.cbook import flatten
from collections import Counter
din={'x':[['a','b','c'],['a','e','d', 'c']], 'y': [['h','j'],['h','j','j']]}
def foo(x):
df = pd.DataFrame()
for a,i in x.items() :
u=pd.DataFrame.from_dict(dict(Counter([*flatten(i)])), orient ='index').reset_index().rename(columns ={'index':a,0:str(a)+'_number'})
df=pd.concat([df,u])
return df
foo(din)
·
Santiago Trujillo
Relatório
Responde à pergunta