Can somebody tell me how to convert list1 to dic_list with all keys equal to the sequence of elements of the list and all values in dictionary equal to the elements in list split by ','?
input:
list1 = ['1,2,3','4,5,6','7,8']
expected output:
dic_list = {0:['1','2','3'],1:['4','5','6'],2:['7','8']}
I created a new list2:
list2 = []
for num in range(0,len(list1)):
list2.append(num)
dic_list = dict(zip(list2,list1))
But my output is:
dic_list = {0:'1,2,3',1:'4,5,6',2:'7,8'}
You can try this:
list1 = ['1,2,3','4,5,6','7,8']
final_list = {i:a.split(',') for i, a in enumerate(list1)}
Output:
{0: ['1', '2', '3'], 1: ['4', '5', '6'], 2: ['7', '8']}
Or, using the builting dict function:
final_list = dict(enumerate(map(lambda x:x.split(','), list1)))
Output:
{0: ['1', '2', '3'], 1: ['4', '5', '6'], 2: ['7', '8']}
You need to split the strings to form lists:
list1 = ['1,2,3', '4,5,6', '7,8']
dic_list = {k: v.split(',') for k, v in enumerate(list1)}
dic_list
output:
{0: ['1', '2', '3'], 1: ['4', '5', '6'], 2: ['7', '8']}
You can enumerate to get the key and split the list with ',' to get desired value.
list1 = ['1,2,3', '4,5,6', '7,8']
Output = {key: value.split(',') for key, value in enumerate(list1)}
Output
Related
I need to delete the elements that are duplicated in a dictionary like this:
{
1: ['1', '2', '3'],
2: ['4', '3', '6', '7'],
3: ['8', '1', '9']
}
as to make the final result like this
{
1: ['1', '2', '3'],
2: ['4', '6', '7'],
3: ['8', '9']
}
Please help me how to do that, I have no idea
Assuming d the input, you can use a set to keep track of the seen values. Here using a dictionary comprehension and "cheating" a bit to add the values:
seen = set()
out = {k: [x for x in v
if x not in seen and not seen.add(x)]
for k,v in d.items()}
Output:
{1: ['1', '2', '3'],
2: ['4', '6', '7'],
3: ['8', '9']}
Same with a classical loop:
out = {}
seen = set()
for k,v in d.items():
l = []
for x in v:
if x not in seen:
seen.add(x)
l.append(x)
out[k] = l
Rehashing the same old seen-set solution is boring :-P. Let's have some fun with Counters:
from collections import Counter
d = {
1: ['1', '2', '3'],
2: ['4', '3', '6', '7'],
3: ['8', '1', '9']
}
seen = Counter()
for a in d.values():
uniq = Counter(dict.fromkeys(a, 1))
a[:] = uniq - seen
seen |= uniq
print(d)
Each list is first deduplicated on its own by using a dict. Then turned into a Counter so we can conveniently subtract the previously seen values. Write the new ones into the list and add them to the seen ones.
Try it online!
You could do the same with set union and difference operators. As sets are unordered the final list would need to be sorted. Again assuming d is the original dictionary.
s = set()
for k in d:
z = d[k]
d[k]= sorted(list(set(z).difference(s)))
s |= set(z)
I'm trying to insert the number "1" to the list in every position with a for loop and eventually get all possible lists in python.
For Example:
l = ["2","3","6"]
number = "1"
output = [["1","2","3","6"],["2","1","3","6"],["2","3","1","6"],["2","3","6","1"]]
l = ["2","3","6"]
list_of_nrs = []
for index in range(len(l)+1):
l.insert(index, "1")
list_of_nrs.append(l)
del l[index]
print(list_of_nrs)
So I've tried it like the code above me, but the output I get is:
[['2', '3', '6'], ['2', '3', '6'], ['2', '3', '6'], ['2', '3', '6']]
It seems like there is a problem between the append and del function.
When you append the output list,you use reference value like pointers in C.Whenever change the value of it in anywhere,it change all over the program.So you have to create new list value.You can use like this:
l = ["2","3","6"]
list_of_nrs = []
for index in range(len(l)+1):
temp = list(l) # temp is a new list now, it wont refer to l anymore
temp.insert(index, "1")
list_of_nrs.append(temp)
print(list_of_nrs)
Your list_of_nrs contains four references to the same list that you keep repeatedly modifying; appending it to the list doesn't create a copy of it!
To create a copy, use .copy():
l = ["2", "3", "6"]
list_of_nrs = []
for index in range(len(l)+1):
l.insert(index, "1")
list_of_nrs.append(l.copy()) # note the .copy()!
del l[index]
print(list_of_nrs)
prints:
[['1', '2', '3', '6'], ['2', '1', '3', '6'], ['2', '3', '1', '6'], ['2', '3', '6', '1']]
The above is the fix that makes the minimal change to your original code; for another approach entirely, you could construct new lists by slicing the original list at different points:
l = ["2", "3", "6"]
list_of_nrs = [
l[:i] + ["1"] + l[i:]
for i in range(len(l)+1)
]
print(list_of_nrs)
I tried to create a dictionary with nested loops but failed. I do not know what's wrong:
dict={}
for i in range(0,4):
node_1=str(i)
for j in range(0,4):
node_2=str(j)
dict[node_1]=[node_2]
print(dict)
It should have created:
{'0':['1','2','3'],'1':['0','2','3'],'2':['0','1','3']}
In your code, you are overwriting the previous j value with the new j value. Instead, you should be appending it to a list.
mydict = {}
for i in range(0,4):
node_1 = str(i)
mydict[node_1] = [] # assign empty list
for j in range(0,4):
node_2 = str(j)
mydict[node_1].append(node_2) # append in list
print(mydict)
Output:
{'0': ['0', '1', '2', '3'], '1': ['0', '1', '2', '3'], '2': ['0', '1', '2', '3'], '3': ['0', '1', '2', '3']}
Note: You should not name your variable dict which is the name for a built-in method.
Something like this?:
d = {}
for i in range(0,4):
node_1=str(i)
for j in range(0,4):
node_2=str(j)
if node_1 not in d:
d[node_1] = []
d[node_1].append(node_2)
print(d)
Please do not use dict for variable name.
I've written a script to make the length of all the lists at least 3 no matter what are their individual length at this moment.
Currently the list of lists I have:
item_list = [['1','2'],['3','4','5'],['2','4','5'],['1']]
I've tried with:
item_list = [['1','2'],['3','4','5'],['2','4','5'],['1']]
for item in item_list:
if len(item)<3:
item.extend([""])
elif len(item)<2:
item.extend([""]*2)
print(item_list)
Output I'm getting:
[['1', '2', ''], ['3', '4', '5'], ['2', '4', '5'], ['1', '']]
Desired output:
[['1', '2', ''], ['3', '4', '5'], ['2', '4', '5'], ['1', '','']]
How can I make the length of all the lists at least 3 irrespective of their current length?
for item in item_list:
item += ['']*(3-len(item))
You have written the order in reverse
item_list = [['1','2'],['3','4','5'],['2','4','5'],['1']]
for item in item_list:
if len(item)<2:
item.extend([""]*2)
elif len(item)<3:
item.extend([""])
print(item_list)
I am trying to practice to make a method.
s = '132'
lis = list(s)
result = []
lisc = lis[:]
for item in lis:
for i in range(1,len(lis)):
lisc.remove(item)
lisc.insert(i,item)
print("lis : ", lisc)
result.append(lisc)
print(result)
The result is :
lis : ['3', '1', '2']
lis : ['3', '2', '1']
lis : ['2', '3', '1']
lis : ['2', '1', '3']
lis : ['1', '2', '3']
lis : ['1', '3', '2']
[['1', '3', '2'], ['1', '3', '2'], ['1', '3', '2'], ['1', '3', '2'], ['1', '3', '2'], ['1', '3', '2']]
I don't get why the result is appending the original lisc instead of the modified lisc in the loop.
I tried result.append(lisc[:]) and it works.
for item in lis:
for i in range(1,len(lis)):
lisc.remove(item)
lisc.insert(i,item)
print("lis : ", lisc)
result.append(lisc[:])
print(result)
Can anyone answer my question?
Thank you in advance.
In the first version of your code, you append the same list that it's being updated at every iteration. In the end, all the lisc you appended will be a reference to the same list
In the second version, you append a new copy of the modified list. Those copies are all different object