CODEWITHSUNDEEP
python
abd
abd

Reputation: 155

How to merge the list elements

Input

['select', '*', 'from', 'ak','.','person']

I need to create a dictionary and merger the element after from

Expected Output

['select', '*', 'from', 'ak.person']

Code is below

m = []
for i in a:
    if '.' == i:
        ind  = a.index('.')
        m.append(a[ind-1] + a[ind] + a[ind+1])
    else:
        m.append(i)

My output >> ['select', '*', 'from', 'ak', 'ak.person', 'person'] Expected is ['select', '*', 'from', 'ak.person']

Upvotes: 1

Views: 76

Answers (5)

Xochozomatli
Xochozomatli

Reputation: 33

Golf answer:

print(a[:(i:=(a:=['select', '*', 'from', 'ak','.','person']).index('from')+1)]+[''.join(a[i:])])

Upvotes: 0

j__carlson
j__carlson

Reputation: 1348

Try this:

def my_func(my_list):
    lp=my_list[(my_list.index('from')+1):(len(my_list))]
    ans=my_list[0:(my_list.index('from')+1)]+["".join(lp)]
    return ans

my_func(['select', '*', 'from', 'ak','.','person'])

Output:

['select', '*', 'from', 'ak.person']

Alternate Solution:

def sub_func(my_list):
    lp=my_list[(my_list.index('.')-1):(my_list.index('.')+2)]
    ans=my_list[0:(my_list.index('.')-1)]+["".join(lp)]+my_list[(my_list.index('.')+2):(len(my_list))]
    return ans

def my_func(my_list):
    lst=my_list
    for i in range(my_list.count('.')):
        lst=(lambda x: sub_func(x))(lst)
    return lst

my_func(['select', '*', 'from', 'ak','.','person','where','foo','.','bar', '=', '30'])

Output:

['select', '*', 'from', 'ak.person', 'where', 'foo.bar', '=', '30']

This solution targets any elements separated by a period and joins them. Unlike the original solution, it will function with lists containing multiple periods and with lists that do not use a 'from' statement.

Upvotes: 0

Manu Manjunath
Manu Manjunath

Reputation: 364

Try this. I first extracted a sub-list after 'from' and merged that.

orig = ['select', '*', 'from', 'ak', '.' ,'person']

from_pos = orig.index('from')
sublist = orig[from_pos+1:from_pos+4]

new_string = ''
for txt in sublist:
    new_string += txt

new_list = orig[0:from_pos+1]
new_list.append(new_string)
print(orig)
print(new_list)

And if you have any where clause or group by after that, you can try this -

orig = ['select', '*', 'from', 'ak', '.' ,'person', 'where', 'filter', 'group', 'by']

from_pos = orig.index('from')
sublist = orig[from_pos+1:from_pos+4]

new_string = ''
for txt in sublist:
    new_string += txt

new_list = orig[0:from_pos+1]
new_list.append(new_string)

where = orig[from_pos+4:]
new_list = new_list + where
print(orig)
print(new_list)

You get - ['select', '*', 'from', 'ak.person', 'where', 'filter', 'group', 'by']

Upvotes: 0

vnk
vnk

Reputation: 1082

The loop here checks if the previous element is 'from' and if so, it joins the following three elements that comes after it.
This should work for the test cases that follow the same pattern like you've given (also including longer queries with where mentioned in the updated question) . 

a = ['select', '*', 'from', 'ak','.','person']
m = []

while i< len(a):
    if a[i-1] == "from":
        m.append("".join(a[i:i+3]))
        i+=3
    else:
        m.append(a[i])
        i+=1

Output

['select', '*', 'from', 'ak.person']

Upvotes: 1

Tranbi
Tranbi

Reputation: 12701

Relatively short:

arr = ['select', '*', 'from', 'ak','.','person']
ind = arr.index('from') + 1
# we join the initial array until 'from' then joining the rest:
print(arr[:ind] + ["".join(arr[ind:])])

Upvotes: 1

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