How to replace entries in an array

Question

I have an array, like:

key = ['*', '(DATE*', '*', '*', '*)', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '(GPE*', '*)', '*', '*', '*', '(DATE)', '*']

I have such an array for which I want to perform task like,

run through the array
once I find the entry starting with '(' but not ending with ')'
replace the next '' entries until we don't find ')' and also replace '*)' with the strip of found entry of starting with '('
and if the entry is within '()' should be just stripped. as for 2nd last element (DATE) to be replaced with DATE only

for E.g. we have 2nd entry '(DATE*' followed by '','','*)' so these entries should be replaced with DATE only

output should be:

key = ['*', 'DATE', 'DATE', 'DATE', 'DATE', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', 'GPE', 'GPE', '*', '*', '*', 'DATE', '*']

sagar singh · Accepted Answer · 2018-04-03 10:37:18Z

3

**Nothing but some regex and while loops**
import re
key = key = ['*', '(DATE*', '*', '*', '*)', '*', '*', '*', '*', '*', '*', '*', '*', '*',
             '*', '*', '*', '*', '*', '*', '*', '*', '*', '(GPE*', '*)', '*', '*', '*', '(DATE)', '*']
val = 0
while val < len(key):
    value = key[val]
    if re.findall(r'\(',value):
        value = re.findall(r'\w+', value)[0]
        while re.findall(r'\)', key[val]) == []:
            key[val] = value
            val += 1
        key[val] = value
    val += 1
print key

output - ['*', 'DATE', 'DATE', 'DATE', 'DATE', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', 'GPE', 'GPE', '*', '*', '*', 'DATE', '*']

answered Apr 3, 2018 at 10:37

sagar singh

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Comments

Vikas Periyadath · Accepted Answer · 2018-04-03 09:58:10Z

1

I know its not much pythonic , anyway you can try this one :

key = ['*', '(DATE*', '*', '*', '*)', '*', '*', '*', '*', '*', '*',
   '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '(GPE*', '*)',
   '*', '*', '*', '(DATE)', '*']

for i in key:
    if i.startswith('(') and not (i.endswith(')')):
        a = key[key.index(i)+1:]
        for j in a:
            if j.endswith(')'):
                a = a[:a.index(j)+1]
                break
        for l in range(key.index(i), key.index(i)+len(a)+1):
            key[l] = i.strip('(').strip('*')
    elif i.startswith('(') and i.endswith(')'):
        key[key.index(i)] = i.strip('(').strip(')')

print(key)

It will give O/P like :

['*', 'DATE', 'DATE', 'DATE', 'DATE', '*', '*', '*', '*', '*', '*', 
 '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', 'GPE', 
'GPE', '*', '*', '*', 'DATE', '*']

answered Apr 3, 2018 at 9:58

Vikas Periyadath

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1 Comment

Vikas Periyadath Over a year ago

Glad to hear :)

karthik reddy · Accepted Answer · 2018-04-03 10:08:18Z

`key = ['*', '(DATE*', '*', '*', '*)', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '(GPE*', '*)', '*', '*', '*', '(DATE)', '*']
outKeys = []
isFound = False
for k in key:
    if k.startswith("(") and k.endswith(")"):
        k = k[k.find("(")+1:k.find(")")]
    elif k.startswith("("):
        k = k[k.find("(")+1:k.find("*")]
        isFound = k
    elif k.endswith(")"):
        k = isFound
        isFound = False
    elif isFound:
        k = isFound
    outKeys.append(k)
print(outKeys)`

This will give u output:

['*', 'DATE', 'DATE', 'DATE', 'DATE', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', 'GPE', 'GPE', '*', '*', '*', 'DATE', '*']

Laurent H. · Accepted Answer · 2018-04-03 10:11:16Z

I suggest you this easily readable solution. I defined another list newKey to avoid modifying a list while iterating over its owm elements :

key = ['*', '(DATE*', '*', '*', '*)', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '(GPE*', '*)', '*', '*', '*', '(DATE)', '*']


newKey = []
next_x = None

for x in key:
    if x.startswith('(') and x.endswith(')'):
        newKey.append(x.strip('()*'))
    elif x.startswith('('):
        newKey.append(x.strip('(*'))
        next_x = x.strip('(*')
    elif x.endswith(')'):
        newKey.append(next_x.strip('*)'))
        next_x = None
    elif next_x is not None:
        newKey.append(next_x)
    else:
        newKey.append(x)  

key = newKey[:]

print(key)

grimek · Accepted Answer · 2018-04-03 10:30:46Z

1

You can use below code:

current_entry = None
for i, k in enumerate(key):
    if k.startswith('(') and k.endswith(')'):
        key[i] = k.strip('(').strip(')')
        continue
    if k.startswith('(') and not k.endswith(')'):
        current_entry = k.strip('(').strip('*')
    if current_entry:
        key[i] = current_entry
    if k.endswith(')'):
        current_entry = None

answered Apr 3, 2018 at 10:30

grimek

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Comments

Vanojx1 · Accepted Answer · 2018-04-03 10:16:58Z

0

Can be done using a simple regex:

string = ' '.join(['*', '(DATE*', '*', '*', '*)', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '*', '(GPE*', '*)', '*', '*', '*', '(DATE)', '*'])
result = re.sub(r'\((.*?)\)', lambda m: ' '.join([m.group(1).replace('*', '').strip()
 for n in range(1 if m.group(0).count('*') == 0 else m.group(0).count('*'))]), string).split(' ')
print(result)

answered Apr 3, 2018 at 10:16

Vanojx1

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