在嵌套列表中查找匹配的元素

Question

我有一個這樣的嵌套列表：

lst = [['one two', 'three', '10'], ['spam eggs', 'spam', '8'],
       ['two three', 'four', '5'], ['foo bar', 'foo', '7'],
       ['three four', 'five', '9']] 

最后一個因素是一種可能性。 我需要找到的元素，其中一個元素的第二個和第三個單詞與另一個元素的第一個和第二個單詞匹配，例如：

['one two', 'three', '10'] match ['two three', 'four', '5'] match  ['three four', 'five', '9']

並制作如下鏈：

one two 10 three 5 four 9 five

我了解第一步必須是元素的tokinization：

lst = ([' '.join(x).split() for x in lst])
for i in lst: 
    print(i)

所以我得到

['one', 'two', 'three', '10']
['spam', 'eggs', 'spam', '8']
['two', 'three', 'four', '4']
['foo', 'bar', 'foo', '7']
['three', 'four', 'five', '9']

下一步應該是對列表的每個元素進行某種迭代搜索，但是我對這樣的搜索的Python實現有些困惑。 任何幫助，將不勝感激。

Answer 1

我建議通過以下方式使用熊貓：

import pandas as pd

lst = [['one two', 'three', '10'], ['spam eggs', 'spam', '8'],
   ['two three', 'four', '5'], ['foo bar', 'foo', '7'],
   ['three four', 'five', '9']]

lst = [' '.join(x).split() for x in lst]

#Create a dataframe and merge using the adequate columns

df = pd.DataFrame(lst)
matchedDF = df.merge(df,how='inner',left_on=[1,2],right_on=[0,1],suffixes=['left','right'])

# remove unneccessary columns
cols=matchedDF.columns.tolist()

matchedDF = matchedDF[cols[2:]]

print(matchedDF)

我得到：

    0left  1left  2left 3left 0right 1right 2right 3right
0   one    two  three    10    two  three   four      5
1   two  three   four     5  three   four   five      9

Answer 2

這也可以：

lst = [['one two', 'three', '10'],['spam eggs', 'spam', '8'], ['two three', 'four', '5'], ['foo bar', 'foo', '7'], ['three four', 'five', '9']] 
lst = ([' '.join(x).split() for x in lst])

match, first = [], True
for i in lst:
    for j in lst:
        if i[0] == j[1] and i[1] == j[2]:
            if first:
                match.append(j)
                first = False
            match.append(i)

for i in match:
    if i == match[len(match)-1]: print(i)
    else: print ("{} match ".format(i), end=' ')

for i in match:
    if i == match[0]: print (i[0], i[1], i[3], end=' ')
    elif i == match[len(match)-1]: print (i[1], i[3], i[2])
    else: print (i[1], i[3], end=' ')

for i in match循環中for i in match的第一個輸出：

['one', 'two', 'three', '10'] match  ['two', 'three', 'four', '5'] match ['three', 'four', 'five', '9']

第二個：

one two 10 three 5 four 9 five

Answer 3

您可以使用itertools

# import itertools
import itertools
# search for the item after generating a chain
item in itertools.chain.from_iterable(lst)

Answer 4

試試這個：

lst = [['one two', 'three', '10'], ['spam eggs', 'spam', '8'],
       ['two three', 'four', '5'], ['foo bar', 'foo', '7'],
       ['three four', 'five', '9']]

lst = [' '.join(x).split() for x in lst]
for i in lst: 
    print(i)

# ---------------------------------------------------------------

st = set()
for i in [set(x) for x in lst]:
    st |= i

print(st)
print(list(st))

輸出：

['one', 'two', 'three', '10']
['spam', 'eggs', 'spam', '8']
['two', 'three', 'four', '5']
['foo', 'bar', 'foo', '7']
['three', 'four', 'five', '9']
{'bar', 'spam', '9', 'one', 'five', 'three', 'two', '8', 'four', '5', 'foo', '10', '7', 'eggs'}
['bar', 'spam', '9', 'one', 'five', 'three', 'two', '8', 'four', '5', 'foo', '10', '7', 'eggs']