df=pd.DataFrame({"a":[1,2,3,[4,5],["apple","pear"]]})
df.replace({[4,5]:4.5})
df.replace({["apple","pear"]:"apple"})
Here I got TypeError. I want to replace specific lists and there is no any regulation between the list which is to be replaced and the object used to replace the list.
This is not a trivial problem, because DataFrames are not designed to work with mutable objects like lists, sets, or dicts.
You can determine the index of match and replace accordingly.
m = [v == [4, 5] for v in df['a']]
df.loc[m, 'a'] = 4.5
df
a
0 1
1 2
2 3
3 4.5
4 [apple, pear]
A similar procedure follows for ['apple', 'pair']. You can form a function from this if you so wish:
def replace(df, col, key, val):
m = [v == key for v in df[col]]
df.loc[m, col] = val
replace(df, 'a', [4, 5], 4.5)
replace(df, 'a', ['apple', 'pear'], 'apple')
df
a
0 1
1 2
2 3
3 4.5
4 apple
Note: The function works in-place.
v == key for v in df[col] is not the correct way to do key matching in Pandas.There is one way using astype , Even it work , but I still highly recommend you using cold's answer.
df.astype(str).replace({'[4, 5]':4.5,"['apple', 'pear']":"apple"})
Out[159]:
a
0 1
1 2
2 3
3 4.5
4 apple
I had a similar problem, since the places I had listed on a column needed to be standardized. First I tried to give a list as key and the standard word as value, which of course failed. So, I made a function to expand the values on the list as keys, and assign the standard word as value for all of them:
def list_to_dict(cities):
new_dict = {}
for key in cities:
value = cities[key]
for item in value:
new_dict[item] = key
return new_dict
With this, I got to clean this list of words aimed to mean Mexico City in Spanish (of course my set is larger and for more places, but this is an illustrative sub-group):
ciudades = list_to_dict({'Ciudad De Mexico' : ['Ciudad De México', 'Cuajimalpa De Morelos', 'Mexicocity', 'Ciudad De Mexico', 'Miguel Hidalgo, Cdmx', 'Df', 'Cmx', 'Ciudad De M', 'Cdmx', 'Ciudad De M?Xico', 'C.D. M.X,', 'Mx-Cdm', 'Cuidad De Mexico', 'Dif', 'D.F.', 'D.F', 'DF', 'Distrito', 'Mexico City', 'Coyoacan', 'Mx-Cdm', 'Cdmex', 'Mx-Dif', 'Mexico Df', 'Ciudad_De_M']}
Resulting in:
