检查来自不同列表的元素是否在 df 列和 append 到另一列

Question

I have a df like this:

Casa	Name	Clase_jfs	Categoria
Just_For_Sports	mochila reebok active	ACCESORIOS	mochila
Just_For_Sports	tubo lejopi de pelotas softee	ACCESORIOS	tubo
Just_For_Sports	pack de medias puma x2	ACCESORIOS	pack
Just_For_Sports	gorro adidas de natación 3 rayas	ACCESORIOS	natacion

And 27 different Lists like these:

MODA=['mochila','wear', 'urban', 'pack']
TENIS=['tubo', 'raqueta','red']
NATACION=['natacion', 'pileta','tapon']

on the other hand I have an empty list:

intermedia1=[]

this is my current script:

for element in df_JFS['Categoria']:       
    if  element in VOLEY:
        intermedia1.append('VOLEY')
    elif element in UNIFORMES:
        intermedia1.append('UNIFORMES')
    elif element in TREKKING_OUTDOOR_ADVENTURE:
        intermedia1.append('TREKKING_OUTDOOR_ADVENTURE')        
    elif element in TRAINING:
        intermedia1.append('TRAINING')        
    elif element in TENIS:
        intermedia1.append('TENIS')        
    elif element in SURF:
        intermedia1.append('SURF')        
    elif element in SQUASH:
        intermedia1.append('SQUASH')  
    elif element in SKATEBOARD:
        intermedia1.append('SKATEBOARD')                    
    elif element in RUNNING:
        intermedia1.append('RUNNING')        
    elif element in RUGBY:
        intermedia1.append('RUGBY')
    elif element in PING_PONG:
        intermedia1.append('PING_PONG')
    elif element in PESAS:
        intermedia1.append('PESAS')
    elif element in PADDLE:
        intermedia1.append('PADDLE')
    elif element in NATACION:
        intermedia1.append('NATACION')
    elif element in MODA:
        intermedia1.append('MODA')
    elif element in INFANTIL:
        intermedia1.append('INFANTIL')
    elif element in HOCKEY:
        intermedia1.append('HOCKEY')
    elif element in HANDBALL:
        intermedia1.append('HANDBALL')
    elif element in GOLF:
        intermedia1.append('GOLF')
    elif element in FUTBOL:
        intermedia1.append('FUTBOL')
    elif element in FRONTON:
        intermedia1.append('FRONTON')
    elif element in CICLISMO:
        intermedia1.append('CICLISMO')
    elif element in BASQUET:
        intermedia1.append('BASQUET')
    elif element in BASICOS:
        intermedia1.append('BASICOS')
    elif element in BASEBALL_SOFTBALL:
        intermedia1.append('BASEBALL_SOFTBALL')
    elif element in ARTES_MARCIALES_Y_BOX:
        intermedia1.append('ARTES_MARCIALES_Y_BOX')
    elif element in AEROBICS_Y_FITNESS:
        intermedia1.append('AEROBICS_Y_FITNESS')
    else:
        intermedia1.append('OTROS')
        
df_JFS['Categoria']=intermedia1

How can it be done efficiently?

output should look like this:

Casa	Name	Clase_jfs	Categoria
Just_For_Sports	mochila reebok active	ACCESORIOS	MODA
Just_For_Sports	tubo lejopi de pelotas softee	ACCESORIOS	TENIS
Just_For_Sports	pack de medias puma x2	ACCESORIOS	MODA
Just_For_Sports	gorro adidas de natación 3 rayas	ACCESORIOS	NATACION

df['Categoria'] value, should be the name of the list where the word was found

Thanks!

Answer 1

Not sure about the time efficiency, but if you want to prevent boilerplate coding, you can use apply function along with a few other steps:

import pandas as pd
# Defining the lists of data(rest of the code)
# .
# .
myDict ={'MODA':MODA, "TENIS":TENIS, "NATACION":NATACION} 
def search(valueToSearch):
  for key, valuesList in myDict.items():
    if valueToSearch in  valuesList:
      return key
  return "Not Found"
df["Categoria"] = df["Categoria"].apply(search)
df

Output

	Casa	Name	Clase_jfs	Categoria
0	Just_For_Sports	mochila reebok active	ACCESORIOS	MODA
1	Just_For_Sports	tubo lejopi de pelotas softee	ACCESORIOS	TENIS
2	Just_For_Sports	pack de medias puma x2	ACCESORIOS	MODA
3	Just_For_Sports	gorro adidas de natación 3 rayas	ACCESORIOS	NATACION

Note that, you should define the myDict as shown above. If you have any other list, you should define them in myDict variable in the same way.

Answer 2

There are few approaches I would suggests

Approach 1

The complexity of finding something in a list is O(n) . it optimise that you can use a set instead which is O(1) .

MODA = set(['mochila', 'wear', 'urban', 'pack'])

Approach 2

If all of the value of all the list is unique, you can create a dict that map values to key. You can just write a loop to map value to key the result should be like below:

{
 'mochila': "MODA",
 'wear': "MODA",
 'urban': "MODA",
 'pack': "MODA",
 ...
}