Removing values from a list in python

问题

I have a large file of names and values on a single line separated by a space:

name1 name2 name3....

Following the long list of names is a list of values corresponding to the names. The values can be 0-4 or na. What I want to do is consolidate the data file and remove all the names and and values when the value is na.

For instance, the final line of name in this file is like so:

namenexttolast nameonemore namethelast 0 na 2

I would like the following output:

namenexttolast namethelast 0 2

How would I do this using Python?

回答1:

I agree with Justin than using zip is a good idea. The problems is how to put the data into two different lists. Here is a proposal that should work ok.

reader = open('input.txt')
writer = open('output.txt', 'w')
names, nums = [], []
row = reader.read().split(' ')
x = len(row)/2
for (a, b) in [(n, v) for n, v in zip(row[:x], row[x:]) if v!='na']:
    names.append(a)
    nums.append(b)
writer.write(' '.join(names))
writer.write(' ')
writer.write(' '.join(nums))
#writer.write(' '.join(names+nums)) is nicer but cause list to be concat

回答2:

Let's say you read the names into one list, then the values into another. Once you have a names and values list, you can do something like:

result = [n for n, v in zip(names, values) if v != 'na']

result is now a list of all names whose value is not "na".

回答3:

s = "name1 name2 name3 v1 na v2"
s = s.split(' ')
names = s[:len(s)/2]
values = s[len(s)/2:]

names_and_values = zip(names, values)
names, values = [], []
[(names.append(n) or values.append(v)) for n, v in names_and_values if v != "na"]
names.extend(values)

print ' '.join(names)

Update

Minor improvement after suggestion from Paul. I'm sure the list comprehension is fairly unpythonic, as it leverages the fact that list.append returns None, so both append expressions will be evaluated and a list of None values will be constructed and immediately thrown away.

回答4:

or say you have a string which you have read from a file. Let's call this string as "s"

words = filter(lambda x: x!="na", s.split())

should give you all the strings except for "na"

edit: the code above obviously doesn't do what you want it to do.

the one below should work though

d = s.split()
keys = d[:len(d)/2]
vals = d[len(d)/2:]
w = " ".join(map(lambda (k,v): (k + " " + v) if v!="na" else "", zip(keys, vals)))
print " ".join([" ".join(w.split()[::2]), " ".join(w.split()[1::2])])

回答5:

strlist = 'namenexttolast nameonemore namethelast 0 na 2'.split()
vals = ('0', '1', '2', '3', '4', 'na')
key_list = [s for s in strlist if s not in vals]
val_list = [s for s in strlist if s in vals]

#print [(key_list[i],v) for i, v in enumerate(val_list) if v != 'na']
filtered_keys = [key_list[i] for i, v in enumerate(val_list) if v != 'na']
filtered_vals = [v for v in val_list if v != 'na']

print filtered_keys + filtered_vals

If you'd rather group the vals, you could create a list of tuples instead (commented out line)

回答6:

Here is a solution that uses just iterators plus a single buffer element, with no calls to len and no other intermediate lists created. (In Python 3, just use map and zip, no need to import imap and izip from itertools.)

from itertools import izip, imap, ifilter

def iterStartingAt(cond, seq):
    it1,it2 = iter(seq),iter(seq)
    while not cond(it1.next()):
        it2.next()
    for item in it2:
        yield item

dataline = "namenexttolast nameonemore namethelast 0 na 2"
datalinelist = dataline.split()

valueset = set("0 1 2 3 4 na".split())

print " ".join(imap(" ".join, 
                    izip(*ifilter(lambda (n,v): v != 'na', 
                                  izip(iter(datalinelist), 
                                       iterStartingAt(lambda s: s in valueset, 
                                                      datalinelist))))))

Prints:

namenexttolast namethelast 0 2

来源：https://stackoverflow.com/questions/3356460/removing-values-from-a-list-in-python