I have a nested list like this:
lst = [['one two', 'three', '10'], ['spam eggs', 'spam', '8'],
['two three', 'four', '5'], ['foo bar', 'foo', '7'],
['three four', 'five', '9']]
The last element is a kind of probability. What I need is to find elements, where second and third words of one element match first and second word of another, for example:
['one two', 'three', '10'] match ['two three', 'four', '5'] match ['three four', 'five', '9']
And make chains like:
one two 10 three 5 four 9 five
I understand that first step must be tokinization of elements:
lst = ([' '.join(x).split() for x in lst])
for i in lst:
print(i)
So I get
['one', 'two', 'three', '10']
['spam', 'eggs', 'spam', '8']
['two', 'three', 'four', '4']
['foo', 'bar', 'foo', '7']
['three', 'four', 'five', '9']
Next step should be some kind of iterative search over each element of the list, but I am a bit stuck with Python realization of such search. Any help would be appreciated.
I would suggest using pandas in the following way:
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)
I get:
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
This works also:
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=' ')
Where the first for i in match
loop outputs:
['one', 'two', 'three', '10'] match ['two', 'three', 'four', '5'] match ['three', 'four', 'five', '9']
And the second:
one two 10 three 5 four 9 five
You can use itertools
# import itertools
import itertools
# search for the item after generating a chain
item in itertools.chain.from_iterable(lst)
Try this one:
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))
Output:
['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']
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