迭代一个 Python 列表并根据另一个列表过滤元素

Question

I have two lists. The first list has all the names.

firstList = ["mike", "mary", "steve", "jane"]

The second list contains the name I want to filter.

filteredList = ["mike", "mary"]

I want to exclude all names in the filtered list so that the final list looks like this:

finalList = ["steve","jane"]

The names are examples and the number of entries will vary and needs to be dynamic.

I started with this but it didn't run:

for b in firstList:
    for s in filteredList:
        finalList = firstList[firstList[b != s]]
        print(finalList)

Thoughts? Any help would be appreciated.

Answer 1

You can use list comprehension to achieve this:

firstList = ["mike", "mary", "steve", "jane"]
filteredList = ["mike", "mary"]

finalList = [name for name in firstList if name not in filteredList]
# where `finalList` will hold: ['steve', 'jane']

Here you are iterating the firstList and checking the presence of each item in filteredList . If not present, consider it for your new list.

You can also use set() to achieve this as:

set(firstList) - set(filteredList)
# returns: set(['steve', 'jane'])

# if it is must for you to obtain the "finalList" as `list`, 
# you can type-cast it as:
#     finalList = list(set(firstList) - set(filteredList))

But via using set, you won't be able to preserve the order of these names in the firstList . And it will hold only unique items ie repetitive items will be removed. For example: if your "firstList" will contain ["mike", "mike", "mary", "steve", "jane"] , finalList will just contain single "mike" whereas in the list comprehension based solution I proposed above, you'll see "mike" twice.

---

In the code you shared, it could be fixed as:

finalList = []

for b in firstList:
    if b not in filteredList: # checking the presence of "b" in "filteredList"
        finalList.append(b)

# where `finalList` will contain: ['steve', 'jane']

Answer 2

Use numpy.setdiff1d

#import numpy as np
np.setdiff1d(firstList, filteredList).tolist()

or set with - operator

list(set(firstList) - set(filteredList))

or filter

list(filter(lambda x: x not in filteredList, firstList))

Output

['steve', 'jane']