如果满足某些条件，则从包含元组列表的列表中删除元组

Question

I have a list containing list of tuples which was obtained after applying postag on word tokenizer. the sample is

    lis=[[[('This', 'DT') ('PM', 'NNP') ('Doctor', 'NNP'), ('Sambit', 'NNP'), ('Patra', 'NNP'), ('Spokesperson', 'NNP')]],[[('Can', 'MD'), ('Media', 'NNP'), ('lambast', 'VB'), ('Sonia', 'NNP'), ('Gandhi', 'NNP'), ('up', 'RP'), ('Dalit', 'NNP'), ('Sitaram', 'NNP'), ('Dalit', 'NNP'), ('President', 'NNP')]]]

I want to remove the tuples when the second element of the tuple is 'NNP'.The OutputList will look like this;

    Final_lis=[[[('This', 'DT')]],[[('Can', 'MD'), ('lambast', 'VB'), ('up', 'RP')]]]

I am writing the code:

   print(len(lis[0][1])) #to print the length of first list containing tuples        
   f_list=[]
   for i in range(0,len(lis)):
       for j in range(len(lis[l])):
           if lis[i][j][1]!='NNP':
              f_list.append(lis[i][j])

But it's showing error

    Traceback (most recent call last):
    File "<ipython-input-51-02562b867f97>", line 1, in <module>
runfile('C:/Users/meet/t1.py', wdir='C:/Users/meet')

    File "C:\Users\meet\Anaconda3\lib\site- 
   packages\spyder\utils\site\sitecustomize.py", line 880, in runfile
       execfile(filename, namespace)

     File "C:\Users\meet\Anaconda3\lib\site- 
   packages\spyder\utils\site\sitecustomize.py", line 102, in execfile
       exec(compile(f.read(), filename, 'exec'), namespace)

     File "C:/Users/meet/t1.py", line 9, in <module>
       print(len(lis[0][1]))

    IndexError: list index out of range

Answer 1

Try this:

flist = []
for j in lis:
    for k in j:
        for l in k:
            if l[1] != 'NNP':
                flist.append(l)
print(flist)

This gives you a single list and not a list of lists. Besides your lis has some issues as there is no , between some tuples.

Answer 2

The error message says that you are trying to access the second (zero indexed) element of the first list of your object, while this list has only one element. Your original code has one loop to few, hence it will always output all the lists due to comparing a list to the string.

The following code should yield the expected output if you want to keep the previous structure (it may maintain empty lists though):

f_list = [
    [
        [
            (val, tag) for val, tag in inner_list if tag != 'NNP'
        ] for inner_list in outer_list
    ]
    for outer_list in lis
]
# result: [[[('This', 'DT')]], [[('Can', 'MD'), ('lambast', 'VB'), ('up', 'RP')]]]

If you only need a list of tuples the following snippet will give the expected outcome:

f_list = [
    (val, tag)
    for outer_list in lis
    for inner_list in outer_list
    for val, tag in inner_list
    if tag != 'NNP'
]
# result: [('This', 'DT'), ('Can', 'MD'), ('lambast', 'VB'), ('up', 'RP')]

Answer 3

这将给您您想要的：

result = [[filter(lambda x: x[1] != 'NNP', a[0])] for a in lis]