So what I was trying to do was output the string "33k22k11k", which is just the last value followed by the reversed last key followed by the second last value followed by the second last reversed key and so on. I'm not sure how to get the reversed key value for the specific loop that I am in. From the code I currently I have, I get the output:
dict = {"k1":1, "k2":2, "k3":3}
current=""
current_k=""
for k,v in dict.items():
for i in k:
current_k=i+current_k
current=str(v)+current_k+current
print(current)
print(current_k)
33k2k1k22k1k11k
3k2k1k
Edited
First of all, if you are on python < 3.6, dict does not keep the order of items. You might want to use collections.OrderedDict for your purpose.
d = {"k1":1, "k2":2, "k3":3}
d.keys()
# dict_keys(['k2', 'k1', 'k3'])
whereas,
d = OrderedDict()
d['k1'] = 1
d['k2'] = 2
d['k3'] = 3
d.keys()
# odict_keys(['k1', 'k2', 'k3'])
With our new d, you can either add the key and values and reverse it:
res = ''
for k, v in d.items():
res += str(k) + str(v)
res[::-1]
# '33k22k11k'
or reversely iterate:
res = ''
for k, v in reversed(d.items()):
res += str(v)[::-1] + str(k)[::-1]
res
# '33k22k11k'
I may be wrong but it seems like you would want to reset the value of current_k each time you access a new key
dict = {"k1":1, "k2":2, "k3":3}
current=""
for k,v in dict.items():
current_k=""
for i in k:
current_k=i+current_k
current=str(v)+current_k+current
print(current)
print(current_k)
Why not simply do:
print(''.join([a+str(b) for a,b in dict.items()])[::-1])
Output:
"33k22k11k"
But if the values are different from the keys, do:
print(''.join([str(b)[::-1]+a for a,b in dict.items()[::-1]]))
You can use the Python map function to create the reversed string(using f-string) for each key/value pair and then join it.
dict1 = {"k1":1, "k2":2, "k3":3}
new_dict = "".join(map(lambda k, v: f'{k}{v}'[::-1] , dict1.keys(), dict1.values()))
Output:
33k22k11k
You can do something like this perhaps:
dict = {"k1":1, "k2":2, "k3":3}
print("".join(list(reversed([str(v)+"".join(reversed(k)) for k, v in dict.items()]))))
Output:
33k22k11k
Related
Is it possible to create a list comprehension for the below for loop?
for key, value in some_dict.items():
if my_value in value:
my_new_var = value['sub_item'].split[0]
This will work :
d={'name':'Yash','age':16}
###############################
keys_values = d.items() # Bonus: Converting the dictionary keys and values to string
d = {str(key): str(value) for key, value in keys_values}#--- ''
#########################
my_value='Yash'
my_new_var = [v for k, v in d.items() if my_value in v] [-1] # edit: now it'll take last value
print(my_new_var)
i managed to do something like this:
out = [value['sub_value'].split[0] for key, value in some_dict.items() for i in value if my_new_var in value]
I have an existing list of Key, Value pairs in my current dictionary called total_list. I want to check my list to see if the length of each Key == 1 in total_list, I want to add that key and its value pair to a new dictionary. This is the code that I've come up with.
total_list = {104370544: [31203.7, 01234], 106813775: [187500.0], 106842625: [60349.8]}
diff_so = defaultdict(list)
for key, val in total_list:
if len(total_list[key]) == 1:
diff_so[key].append[val]
total_list.pop[key]
But I keep getting an error with
"cannot unpack non-iterable int object".
I was wondering if there's anyway for me to fix this code for it to run properly?
Assuming that the OP means a string of one character by length = 1 of the key.
You can do this:
total_list = [{'abc':"1", 'bg':"7", 'a':"7"}]
new_dict = {}
for i in total_list:
for k,v in i.items():
if len(k) == 1:
new_dict[str(k)] = v
else:
pass
print(new_dict)
Output:
{'a': '7'}
After edit:
total_list = {104370544: [31203.7, 1234], 106813775: [187500.0], 106842625: [60349.8]}
new_dict = {}
for k,v in total_list.items():
if len(v) == 1:
new_dict[k] = v
else:
pass
Output:
{'106842625': [60349.8], '106813775': [187500.0]}
You just need a dictionary comprehension
diff_so = {k: v for k, v in total_list.items() if len(v) == 1}
I have a dict as below
{"low":[18,12,9],"medium":[6,3],"high":[2,1],"final":[0]}
and I want to search for a number in this dict and get its respective 'key'
Eg: for 12, i need to return 'low'. slly for 2, return 'high'
You can use a dictionary comprehension for this.
dict = {"low":[18,12,9],"medium":[6,3],"high":[2,1],"final":[0]}
key = {k:v for k, v in dict.items() if 12 in v}
Output
In[1]: key.popitem()[0]
Out[1] : 12
This is a job for next.
my_d = {"low":[18,12,9],"medium":[6,3],"high":[2,1],"final":[0]}
target = 12
res = next((k for k, v in my_d.items() if target in v), 'N\A')
print(res) # low
Note that if your target value exists in more than one keys, this code will return one of them at random1. If that might by the case and depending on the problem you are working on it may be wiser to get all matching keys instead. To do that, use:
res = [k for k, v in my_d.items() if target in v]
1Actually more like in an uncontrolled fashion.
def getKey(number):
for key, value in d.iteritems():
if number in value:
return key
dict = {"low":[18,12,9],"medium":[6,3],"high":[2,1],"final":[0]}
def search_key(val):
for key, value in dict.iteritems():
for i in value:
if i == val:
print "the key is:"+key
return key
#pass any value for which you want to get the key
search_key(9)
I have a dict like this:
d = {'first':'', 'second':'', 'third':'value', 'fourth':''}
and I want to find first non-empty value (and it's name, in this example 'third'). There may be more than one non-empty value, but I only want the first one I find.
How can I do this?
Use an OrderedDict which preserves the order of elements. Then loop over them and find the first that isn't empty:
from collections import OrderedDict
d = OrderedDict()
# fill d
for key, value in d.items():
if value:
print(key, " is not empty!")
You could use next (dictionaries are unordered - this somewhat changed in Python 3.6 but that's only an implementation detail currently) to get one "not-empty" key-value pair:
>>> next((k, v) for k, v in d.items() if v)
('third', 'value')
Like this?
def none_empty_finder(dict):
for e in dict:
if dict[e] != '':
return [e,dict[e]]
d = {'first':'', 'second':'', 'third':'value', 'fourth':''}
for k, v in d.items():
if v!='':
return k, v
Edit 1
from the comment if the value is None or '' we better use if v: instead of if v!=''. if v!='' only check the '' and skip others
You can find empty elements and make a list of them:
non_empty_list = [(k,v) for k,v in a.items() if v]
By using list comprehension, you can list all the non-empty values and then fetch the 0th value:
[val for key, val in d.items() if val][0]
I have a set of reactions (keys) with values (0.0 or 100) stored in mydict.
Now I want to place non zero values in a new dictionary (nonzerodict).
def nonzero(cmod):
mydict = cmod.getReactionValues()
nonzerodict = {}
for key in mydict:
if mydict.values() != float(0):
nonzerodict[nz] = mydict.values
print nz
Unfortunately this is not working.
My questions:
Am I iterating over a dictionary correctly?
Am I adding items to the new dictionary correctly?
You are testing if the list of values is not equal to float(0). Test each value instead, using the key to retrieve it:
if mydict[key] != 0:
nonzerodict[key] = mydict[key]
You are iterating over the keys correctly, but you could also iterate over the key-value pairs:
for key, value in mydict.iteritems():
if value != 0:
nonzerodict[key] = value
Note that with floating point values, chances are you'll have very small values, close to zero, that you may want to filter out too. If so, test if the value is close to zero instead:
if abs(value) > 1e-9:
You can do the whole thing in a single dictionary expression:
def nonzero(cmod):
return {k: v for k, v in cmod.getReactionValues().iteritems() if abs(v) > 1e-9}
Its simple and you can it by below way -
>>> d = {'a':4,'b':2, 'c':0}
>>> dict((k,v) for k,v in d.iteritems() if v!=0)
{'a': 4, 'b': 2}
>>>
Replace if condition in you code with:
if mydict[key]:
nonzerodict[key] = mydict[key]
Your solution can be further simplified as:
def nonzero(cmod):
mydict = cmod.getReactionValues()
nonzerodict = {key: value for key, value in mydict.iteritems() if value}