Append duplicate items at the end of list whithout changing the order
Question:
I am new to python and was trying to Append duplicate items at the end of list whithout changing the order
testlist = [1, 2, 32, 8, 1, 17, 5, 2, 42, 13, 56]
def duplicate(alist):
p = len(alist)
duplicate = False
for i in range(0, p):
for j in range (i + 1, p):
if alist[i] == alist[j]:
b = alist.index(alist[j])
a = alist.pop(b)
alist.append(a)
p -= 1
duplicate = True
print alist
if duplicate == False:
print "No duplicate item found"
duplicate(testlist)
OUTPUT : [32, 8, 1, 17, 5, 2, 42, 13, 56, 1, 2]
DESIRED OUTPUT : [1, 2, 32, 8, 17, 5, 42, 13, 56, 1, 2]
Any help what wrong I am doing here
Answers:
I think that in this case the creation of a new list is more efficient and clear in comparison with the permutations in the original list:
testlist = [1, 2, 32, 8, 1, 17, 5, 2, 42, 13, 56]
def duplicate(alist):
filtered, duplicates = [], []
for element in alist:
if element in filtered:
duplicates.append(element)
continue
filtered.append(element)
if not duplicates:
print "No duplicate item found"
return alist
return filtered + duplicates
new_list = duplicate(testlist)
print new_list
You may use the Collections module to get OrderedDict for maintaining the order of elements.
The technique we use here is to create a dictionary to store the number of occurrences of each element in the array and use the dict to lookup the number of occurrences for later use.
In the for loop we look up if there is already element present in the dict using the get method. If true then we increase the counter else initialize the counter to be zero.
import collections
lst = [1, 2, 32, 8, 1, 17, 5, 2, 42, 13, 56]
# Create a dictionary storing the the number of occurences
occurences_dict = collections.OrderedDict()
for i in lst:
occurences_dict[i] = occurences_dict.get(i, 0) + 1
final = occurences_dict.keys() + [k for k, v in occurences_dict.items() if v>1]
print final
>>> [1, 2, 32, 8, 17, 5, 42, 13, 56, 1, 2]
I solved it this way. I also made some changes to make the code a bit more Pythonic.
test_list = [1, 2, 32, 8, 1, 17, 5, 2, 42, 13, 56]
def duplicate(a_list):
list_length = len(a_list)
duplicate = False
checked = []
for i in range(list_length):
if a_list.count(a_list[i]) > 1:
if a_list[i] not in checked:
duplicate = True
a_list.append(a_list[i])
checked.append(a_list[i])
if duplicate == False:
print("No duplicate item found")
return None
return a_list
print(duplicate(test_list))
Instead of checking values, compare on index.
Please check this code:
testlist = [1, 2, 32, 8, 1, 17, 5, 2, 42, 13, 56,32]
print("Original list - ",testlist)
tmp_list = []
exclude =[]
for i in range(len(testlist)):
if i == testlist.index(testlist[i]):
tmp_list.append(testlist[i])
else:
exclude.append(testlist[i])
tmp_list.extend(exclude)
testlist = tmp_list
print("Updated list - ",testlist)
Output:
C:Usersdinesh_pundkarDesktop>python c.py
Original list - [1, 2, 32, 8, 1, 17, 5, 2, 42, 13, 56, 32]
Updated list - [1, 2, 32, 8, 17, 5, 42, 13, 56, 1, 2, 32]
C:Usersdinesh_pundkarDesktop>
I have developed a algo which extends the duplicates in such a way that if duplicate contains duplicate we are extends them to end also
eg – input – [1,1,2,2,3,3,4,5,2,5]
output – [1, 2, 3, 4, 5, 2, 3, 2] – (Solution 2)
others ouput – [1, 2, 3, 4, 5, 2, 2, 3] where 2 repeats (Solution 1)
Solution 1 –
def remove_extend(l):
stack = []
duplicates = []
for i in l:
if i not in stack:
stack.append(i)
elif i in stack:
duplicates.append(i)
stack.extend(duplicates)
return stack
l = [1,2,2,2,3,3,4,5]
ans = remove_extend(l)
print(l)
print(ans)
Solution 2
def check_dict(d):
if d == {}:
return False
# stack = []
for k,v in d.items():
if v == 0:
del d[k]
return True
def add_to_stack(stack, d):
for k,v in d.items():
if v>=1:
stack.append(k)
d[k] -= 1
return [stack,d]
def duplicates_extend2(ls):
d = {}
stack = []
for i in ls:
if i in d:
d[i] += 1
else:
d[i] = 1
print(d)
if_dict = check_dict(d)
while if_dict:
ans = add_to_stack(stack, d)
stack = ans[0]
d = ans[1]
if_dict = check_dict(d)
return stack
ls = [1,1,2,2,3,3,4,5,2,5]
ans = duplicates_extend2(ls)
print(ans)
I am new to python and was trying to Append duplicate items at the end of list whithout changing the order
testlist = [1, 2, 32, 8, 1, 17, 5, 2, 42, 13, 56]
def duplicate(alist):
p = len(alist)
duplicate = False
for i in range(0, p):
for j in range (i + 1, p):
if alist[i] == alist[j]:
b = alist.index(alist[j])
a = alist.pop(b)
alist.append(a)
p -= 1
duplicate = True
print alist
if duplicate == False:
print "No duplicate item found"
duplicate(testlist)
OUTPUT : [32, 8, 1, 17, 5, 2, 42, 13, 56, 1, 2]
DESIRED OUTPUT : [1, 2, 32, 8, 17, 5, 42, 13, 56, 1, 2]
Any help what wrong I am doing here
I think that in this case the creation of a new list is more efficient and clear in comparison with the permutations in the original list:
testlist = [1, 2, 32, 8, 1, 17, 5, 2, 42, 13, 56]
def duplicate(alist):
filtered, duplicates = [], []
for element in alist:
if element in filtered:
duplicates.append(element)
continue
filtered.append(element)
if not duplicates:
print "No duplicate item found"
return alist
return filtered + duplicates
new_list = duplicate(testlist)
print new_list
You may use the Collections module to get OrderedDict for maintaining the order of elements.
The technique we use here is to create a dictionary to store the number of occurrences of each element in the array and use the dict to lookup the number of occurrences for later use.
In the for loop we look up if there is already element present in the dict using the get method. If true then we increase the counter else initialize the counter to be zero.
import collections
lst = [1, 2, 32, 8, 1, 17, 5, 2, 42, 13, 56]
# Create a dictionary storing the the number of occurences
occurences_dict = collections.OrderedDict()
for i in lst:
occurences_dict[i] = occurences_dict.get(i, 0) + 1
final = occurences_dict.keys() + [k for k, v in occurences_dict.items() if v>1]
print final
>>> [1, 2, 32, 8, 17, 5, 42, 13, 56, 1, 2]
I solved it this way. I also made some changes to make the code a bit more Pythonic.
test_list = [1, 2, 32, 8, 1, 17, 5, 2, 42, 13, 56]
def duplicate(a_list):
list_length = len(a_list)
duplicate = False
checked = []
for i in range(list_length):
if a_list.count(a_list[i]) > 1:
if a_list[i] not in checked:
duplicate = True
a_list.append(a_list[i])
checked.append(a_list[i])
if duplicate == False:
print("No duplicate item found")
return None
return a_list
print(duplicate(test_list))
Instead of checking values, compare on index.
Please check this code:
testlist = [1, 2, 32, 8, 1, 17, 5, 2, 42, 13, 56,32]
print("Original list - ",testlist)
tmp_list = []
exclude =[]
for i in range(len(testlist)):
if i == testlist.index(testlist[i]):
tmp_list.append(testlist[i])
else:
exclude.append(testlist[i])
tmp_list.extend(exclude)
testlist = tmp_list
print("Updated list - ",testlist)
Output:
C:Usersdinesh_pundkarDesktop>python c.py
Original list - [1, 2, 32, 8, 1, 17, 5, 2, 42, 13, 56, 32]
Updated list - [1, 2, 32, 8, 17, 5, 42, 13, 56, 1, 2, 32]
C:Usersdinesh_pundkarDesktop>
I have developed a algo which extends the duplicates in such a way that if duplicate contains duplicate we are extends them to end also
eg – input – [1,1,2,2,3,3,4,5,2,5]
output – [1, 2, 3, 4, 5, 2, 3, 2] – (Solution 2)
others ouput – [1, 2, 3, 4, 5, 2, 2, 3] where 2 repeats (Solution 1)
Solution 1 –
def remove_extend(l):
stack = []
duplicates = []
for i in l:
if i not in stack:
stack.append(i)
elif i in stack:
duplicates.append(i)
stack.extend(duplicates)
return stack
l = [1,2,2,2,3,3,4,5]
ans = remove_extend(l)
print(l)
print(ans)
Solution 2
def check_dict(d):
if d == {}:
return False
# stack = []
for k,v in d.items():
if v == 0:
del d[k]
return True
def add_to_stack(stack, d):
for k,v in d.items():
if v>=1:
stack.append(k)
d[k] -= 1
return [stack,d]
def duplicates_extend2(ls):
d = {}
stack = []
for i in ls:
if i in d:
d[i] += 1
else:
d[i] = 1
print(d)
if_dict = check_dict(d)
while if_dict:
ans = add_to_stack(stack, d)
stack = ans[0]
d = ans[1]
if_dict = check_dict(d)
return stack
ls = [1,1,2,2,3,3,4,5,2,5]
ans = duplicates_extend2(ls)
print(ans)