i got this code, and i want to add the print in "lista_3"
how can i do it? Thanks!
lista_1 = ["h",'o','l','a',' ', 'm','u','n','d','o']
lista_2 = ["h",'o','l','a',' ', 'l','u','n','a']
lista_3 = []
set_1 = set(lista_1)
set_2 = set(lista_2)
print(set_1.intersection(set_2))
Related
text = "hellovision hey creator yoyo b creator great publisher"
I want to extract creator's name and publisher's name from text.
The result will be,
creator = hellovision hey, yoyo
publisher = great
How can I get text using regular expression?
Do I need to use span()..?
This is my code.
def preprocess2(text):
text_list = test.split(' ')
lyricist = []
composer = []
music_arranger = []
temp = []
lyricist.clear()
composer.clear()
music_arranger.clear()
for i in range(0, len(text_list)):
if text_list[i] == 'creator':
print(len(text_list))
for a in range(0, i-1):
temp.append(text_list[a])
lyricist.append(''.join(temp))
temp.clear()
for b in range(0, i+1):
print(b)
text_list.pop(b)
print(len(text_list))
break
elif text_list[i] == 'pulisher':
for a in range(0, i-1):
temp.append(text_list[a])
composer.append(''.join(temp))
temp.clear()
for b in range(0, i+1):
text_list.pop(b)
break
i = i +1
return text_list
If you split your array using regex with a capture group, the value that you split on will also be passed into the output array.
You can then loop through looking for 'creator' or 'publisher' and in each case, pass the previous entry into the proper collection.
const text = "hellovision hey creator yoyo b creator great publisher"
const splitArr = text.split(/(creator|publisher)/)
const creators = [], publishers = []
let i = -1, len = splitArr.length
while(++i < len){
if(splitArr[i] == "creator") creators.push(splitArr[i-1].trim())
else if(splitArr[i] == "publisher") publishers.push(splitArr[i-1].trim())
}
console.log("creators: ", creators)
console.log("publishers: ", publishers)
What can I do when I get the following error in this code?
def reverse_word(word):
index = len(word)
new_word = []
for i in range(index - 1, -1, -1):
new_word.append(word[i])
return ''.join(new_word)
def reverse_sentence(sentence):
l = sentence.split()
for i in l:
l[i] = reverse_word(i)
l = ' '.join(l)
print(l)
a = "Hello !Nhoj Want to have lunch?"
reverse_sentence(a)
TypeError: list indices must be integers or slices, not str.
What can I write instead of this line:
l[i] = reverse_word(i)
... this line: l[i] = reverse_word(i) throws an error because i is a string (word), but in l[i], i must be an index.
You probably wanted to do something like this:
words = sentence.plit()
new_sentence = []
for i,word in enumerate(words): #or just for word in words, you don't need the index
new_sentence.append(reverse_word(word))
and then join at return time return ' '.join(new_sentence)
This implementation follows your logic but uses strings instead of lists.
def reverse_word(word):
new_word = ''
for i in range(len(word) - 1, -1, -1):
new_word += word[i]
return new_word
def reverse_sentence(sentence):
r = ''
for word in sentence.split():
r += reverse_word(word) + ' '
return r[:-1]
>>> a = "Hello !Nhoj Want to have lunch?"
>>> reverse_sentence(a)
>>> 'olleH johN! tnaW ot evah ?hcnul'
I am having trouble storing the ID to keys, like a sub (parent-child) kind of thing. I spent hours on it and could not figure a way to accomplish this. What output I am expecting is at the end of this post. Any help would be great.
import sys
import collections
dict = collections.OrderedDict()
dict["A.1"] = {"parent_child":0}
dict["A.1.1"] = {"parent_child":1}
dict["A.1.1.1"] = {"parent_child":2}
dict["A.1.1.2"] = {"parent_child":2}
dict["A.1.1.3"] = {"parent_child":2}
dict["A.1.2"] = {"parent_child":1}
dict["A.1.2.1"] = {"parent_child":2}
dict["A.1.2.2"] = {"parent_child":2}
dict["A.1.2.2.1"] = {"parent_child":3}
dict["A.1.2.2.2"] = {"parent_child":3}
dict["A.1.2.3"] = {"parent_child":2}
dict["A.1.3"] = {"parent_child":1}
dict["A.1.4"] = {"parent_child":1}
print(dict)
new_dict = {}
p = 0 # previous index
i = 0 # current
n = 1 # next index
current_PC = 0 # current parent_child
next_PC = 0 # next parent_child
previous_id = ""
current_id = ""
next_id = ""
change_current = True
change = True
lst = []
while(True):
if change_current:
current_id = dict.keys()[i]
current_PC = dict.values()[i]["parent_child"]
change_current = False
try:
next_id = dict.keys()[n]
next_PC = dict.values()[n]["parent_child"]
except:
pass # it will go out of index
print("KEY {0}".format(current_id))
if next_PC > current_PC:
if next_PC - current_PC == 1:
lst.append(next_PC)
next_PC += 1
print("next_PC: {0}".format(next_PC))
if next_PC == current_PC:
new_dict[current_id] = lst
lst = []
break
print(new_dict)
Trying to make output looks like this (at in similar way), the new_dict should look like:
new_dict["A.1"] = ["A.1.1", "A.1.2", "A.1.3", "A.1.4"]
new_dict["A.1.1"] = ["A.1.1.1", "A.1.1.2", "A.1.1.3"]
new_dict["A.1.1.1"] = []
new_dict["A.1.1.2"] = []
new_dict["A.1.1.3"] = []
new_dict["A.1.2"] = ["A.1.2.1", "A.1.2.2", "A.1.2.3"]
new_dict["A.1.2.1"] = []
new_dict["A.1.2.2"] = ["A.1.2.2.1", "A.1.2.2.2"]
new_dict["A.1.2.2.1"] = []
new_dict["A.1.2.2.2"] = []
new_dict["A.1.2.3"] = []
new_dict["A.1.3"] = []
new_dict["A.1.4"] = []
This gives you the output you are asking for. Since i did not see a {"parent_child":...} in you desired output i did not proceed with anything else.
options = ["A.1","A.1.1","A.1.1.1","A.1.1.2","A.1.1.3","A.1.2","A.1.2.1","A.1.2.2","A.1.2.2.1","A.1.2.2.2","A.1.2.3","A.1.3","A.1.4"]
new_dict = {}
for i, key in enumerate(options):
new_dict[key] = []
ls = []
for j, opt in enumerate(options):
if (key in opt) and (len(opt)-len(key)==2):
new_dict[key].append(opt)
print(new_dict)
EDIT
Using the comment of #Ranbir Aulakh
options = ["A.1","A.1.1","A.1.1.1","A.1.1.2","A.1.1.3","A.1.2","A.1.2.1","A.1.2.2","A.1.2.2.1","A.1.2.2.2","A.1.2.3","A.1.3","A.1.4"]
new_dict = {}
for i, key in enumerate(options):
new_dict[key] = []
ls = []
for j, opt in enumerate(options):
if (key in opt) and (len(opt.split("."))-len(key.split("."))==1):#(len(opt)-len(key)==2):
new_dict[key].append(opt)
print(new_dict)
I am getting the error need more than 1 value to unpack i am using defaultDict to make my list to dictionary.
def getTabsCols(request):
proj_name = request.GET.get('projname')
cursor = connection.cursor()
proj_details = TProjects.objects.get(
attr_project_name=proj_name, attr_project_type='Structure', attr_is_active=1)
pid = proj_details.project_id
query = 'call SP_Get_TABCOL_NAMES('+str(pid)+')'
cursor.execute(query)
proj_details = TProjects.objects.get(
attr_project_name=proj_name, attr_project_type='Structure', attr_is_active=1)
pid = proj_details.project_id
query = 'call SP_Get_TABCOL_NAMES('+str(pid)+')'
cursor.execute(query)
result = cursor.fetchall()
tab_list = []
tab_col_list = []
prd = {}
res = []
for row in result:
tabs = collections.OrderedDict()
schema_name = row[1]
table_name = row[3]
tab = (schema_name + '.' + table_name).encode('utf8')
tabs[tab] = row[5].encode('utf8')
tab_list.append(tabs)
d = collections.defaultdict(set)
for k, v in tab_list:
d1[k].append(v)
d = dict((k, tuple(v)) for k, v in d1.iteritems())
return HttpResponse(d, content_type="text/html")
Does anyone know how I get Django Filter build an OR statement? I'm not sure if I have to use the Q object or not, I thought I need some type of OR pipe but this doesn't seem right:
filter_mfr_method = request.GET.getlist('filter_mfr_method')
for m in filter_mfr_method:
designs = designs.filter(Q(mfr_method = m) | m if m else '')
# Or should I do it this way?
#designs = designs.filter(mfr_method = m | m if m else '')
I would like this to be:
SELECT * FROM table WHERE mfr_method = 1 OR mfr_method = 2 OR mfr_method = 3
EDIT: Here is what Worked
filter_mfr_method = request.GET.getlist('filter_mfr_method')
list = []
for m in filter_mfr_method:
list.append(Q(mfr_method = m))
designs = designs.filter(reduce(operator.or_, list))
What about:
import operator
filter_mfr_method = request.GET.getlist('filter_mfr_method')
filter_params = reduce(operator.or_, filter_mfr_method, Q())
designs = designs.filter(filter_params)
Something I used before:
qry = None
for val in request.GET.getlist('filter_mfr_method'):
v = {'mfr_method': val}
q = Q(**v)
if qry:
qry = qry | q
else:
qry = q
designs = designs.filter(qry)
That is taken from one of my query builders.