Question

我有清单：

print (L)
[('bar', 'one'), ('bar', 'two'), ('baz', 'one'), 
 ('baz', 'two'), ('foo', 'one'), ('qux', 'one'), 
 ('qux', 'two'), ('oof', 'two'), ('oof', 'one'), ('oof', 'three')]

我希望按元组中的第一个元素进行分组，并过滤包含one和two作为第二个元素的所有元组。

因此，需要过滤掉('oof', 'two')和('foo', 'one')，因为foo只有一个元素，oof只有3个元素。

预期输出 - 每个第一个元素bar，baz秒为one和two，长度为2：

print(L1)   
[('bar', 'one'), ('bar', 'two'), 
 ('baz', 'one'), ('baz', 'two'), 
 ('qux', 'one'), ('qux', 'two')]

我试试：

L = [b in ['one','two'] for a,b in L]
print (L)
[True, True, True, True, True, True, True, True]

它的优点/ pythonic解决方案是什么？

Answer 1

以下是使用groupby的解决方案：

import itertools, operator

# group the tuples by the first element
result = itertools.groupby(sorted(L), key=operator.itemgetter(0))
# convert the groups to lists
result = [list(group) for _, group in result]
# filter out those lists that don't contain exactly "one" and "two"
result = [group for group in result if set(y for x, y in group) == {'one', 'two'}]
# flatten the nested list into a list of tuples
result = [x for group in result for x in group]

print(result)

请注意，这并不关心重复的元组：

L = [('bar', 'one'), ('bar', 'two'), ('bar', 'two')]
# result = [('bar', 'one'), ('bar', 'two'), ('bar', 'two')]

如果你在输出中不想要这些，你可以像这样重写过滤条件（第二列表理解）：

result = [group for group in result if
             set(y for x, y in group) == {'one', 'two'} and len(group) == 2]

Answer 2

你可以通过pandas groupby ie

完成这项工作

L = [('bar', 'one'), ('bar', 'two'), ('baz', 'one'), 
 ('baz', 'two'), ('foo', 'one'), ('qux', 'one'), 
 ('qux', 'two'), ('oof', 'two'), ('oof', 'one'), ('oof', 'three'),
 ('new','five'),('new','six')]

df  = pd.DataFrame(L)

s = df.groupby(0).size()
temp = s[s==2].index

idx = df[df[0].isin(temp)].groupby(0)[1].apply(lambda x : all(x.isin(['one','two'])))

df[df[0].isin(idx[idx].index)].apply(tuple,1).tolist()

[('bar', 'one'),
 ('bar', 'two'),
 ('baz', 'one'),
 ('baz', 'two'),
 ('qux', 'one'),
 ('qux', 'two')]

Answer 3

这一行怎么样？

data=[('bar', 'one'), ('bar', 'two'), ('baz', 'one'),
 ('baz', 'two'), ('foo', 'one'), ('qux', 'one'),
 ('qux', 'two'), ('oof', 'two'), ('oof', 'two')]

from itertools import groupby

print(list(filter(lambda x:len(x)==2 and sorted((x[1][1],x[0][1]))==['one','two'],[list(b) for a,b in groupby(data,key=lambda x:x[0])])))

输出：

[[('bar', 'one'), ('bar', 'two')], [('baz', 'one'), ('baz', 'two')], [('qux', 'one'), ('qux', 'two')]]

详细：

data=[('bar', 'one'), ('bar', 'two'), ('baz', 'one'),
 ('baz', 'two'), ('foo', 'one'), ('qux', 'one'),
 ('qux', 'two'), ('oof', 'two'), ('oof', 'one'), ('oof', 'three')]

dublicates={}

for i in data:
    if i[0] not in dublicates:
        dublicates[i[0]]=[i[1]]
    else:
        dublicates[i[0]].append(i[1])
print(dublicates)


final=[]
for j,i in dublicates.items():
    if len(i)==2:
        if 'one' and 'two' in i:
            final.extend([(j,'one'),(j, 'two')])

print(final)

输出：

[('baz', 'one'), ('baz', 'two'), ('qux', 'one'), ('qux', 'two'), ('bar', 'one'), ('bar', 'two')]

Answer 4

内置sorted会自动执行此操作。在对元组列表进行排序时，它将按第一项排序，然后按第二项排序，依此类推。

from pprint import pprint
def is_interesting(element):
    a, b = element
    return b in ('one', 'two')

result = sorted(filter(is_interesting, some_list))
pprint(result)

输出为

[('bar', 'one'),
 ('bar', 'two'),
 ('baz', 'one'),
 ('baz', 'two'),
 ('foo', 'one'),
 ('oof', 'two'),
 ('qux', 'one'),
 ('qux', 'two')]

筛选元组列表

4 个答案: