Remove all individual row values that are not in Table2 from Table1, while keeping the count of Tabl-CodePudding

I have two tables:
Table1

Values     | Count  
a, b, e, f | 20 
a, f, j    | 30
b          | 40
c          | 50

Table2

Values     | Count  
a          | 5 
b          | 10
c          | 10
d          | 20

The problem is that I want to remove all individual values that are not in Table2 from Table1, while keeping the count of Table1. For example, because the values, e, f, j are not in Table2, I want to remove only those values from Table1 without changing the count. So the final Table1 needs to look like this.

Table1

Values     | Count  
a, b       | 20 
a,         | 30
b          | 40
c          | 50

This what I've tried so far:
I've converted the both, the Values and Count columns to lists. Split the lists. And then Flattened the lists.

Table1["Values"].tolist()

newList = []
for item in values2:
    newList.append(item.split(","))
    newList

output = [] 
def removeNestings(newList):
    for i in newList:
        if type(i) == list:
            removeNestings(i)
        else:
            output.append(i)

But in the process, I lose the count. That's where I'm stuck. I cannot keep the count. Can someone please help me out?

CodePudding user response：

Try this,

t1 = [["a, b, e, f",20],["a, f, j",30],["b",40],["c",50]]
t2 = [["a",5],["b",10],["c",10],["d",20]]

idx = [i[0] for i in t2]              #["a", "b", "c", "d"]
for item in t1:
    new_value = item[0].split(", ")   #["a", "b", "e", "f"]
    for v in item[0].split(", "):
        if v not in idx:
            new_value.remove(v)       #["a", "b"]
    item[0] = ", ".join(new_value)    #"a, b"
print(*t1, sep="\n")

Result:

[["a, b", 20], ["a", 30], ["b", 40], ["c", 50]]