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I have a data set as below
tmp_dict = {
'a': ?,
'b': ?,
'c': ?,
}
and I have a data is a list of dictionaries like
tmp_list = [tmp_dict1, tmp_dict2, tmp_dict3....]
and I found some of dictionaries are not perfectly have keys about 'a','b','c'.
How do I check and fill the key is not existing
You could try something like this:
# List of keys to look for in each dictionary
dict_keys = ['a','b','c']
# Generate the dictionaries for demonstration purposes only
tmp_dict1 = {'a':[1,2,3], 'b':[4,5,6]}
tmp_dict2 = {'a':[7,8,9], 'b':[10,11,12], 'c':[13,14,15]}
tmp_dict3 = {'a':[16,17,18], 'c':[19,20,21]}
# Add the dictionaries to a list as per OP instructions
tmp_list = [tmp_dict1, tmp_dict2, tmp_dict3]
#--------------------------------------------------------
# Check for missing keys in each dict.
# Print the dict name and keys missing.
# -------------------------------------------------------
for i, dct in enumerate(tmp_list, start=1):
for k in dict_keys:
if dct.get(k) == None:
print(f"tmp_dict{i} is missing key:", k)
OUTPUT:
tmp_dict1 is missing key: c
tmp_dict3 is missing key: b
I think you want this.
tmp_dict = {'a':1, 'b': 2, 'c':3}
default_keys = tmp_dict.keys()
tmp_list = [{'a': 1}, {'b': 2,}, {'c': 3}]
for t in tmp_list:
current_dict = t.keys()
if default_keys - current_dict:
t.update({diff: None for diff in list(default_keys-current_dict)})
print(tmp_list)
Output:
[{'a': 1, 'c': None, 'b': None}, {'b': 2, 'a': None, 'c': None}, {'c': 3, 'a': None, 'b': None}]
You can compare the keys in the dictionary with a set containing all the expected keys.
for d in tmp_list:
if set(d) != {'a', 'b', 'c'}:
print(d)
I have a list of dictionaries, that have same keys and some have different values for those keys. I am trying to append the dictionaries that have different values from the list to keep track of the different values and I would concatenate the values of other keys. For example, I am storing 'a' keys with same values and concatenating the 'b' values that have same 'a':'1'
input list: d = [{'a': '1', 'b': '3'}, {'a': '2', 'b': '4'}, {'a': '1', 'b':'5'}]
output list: p = [{'a':'1', 'b': '35'}, {'a': '2', 'b': '4'}]
So far, I tried the following code, but it doesnt recognize the different values
length = len(p)
j = 0
for i in d:
while j < length:
if p[j]['a'] is not i['a']:
p.append({'a', p[j]['a']})
else:
p[j]['b'] += i['b']
j += 1
j = 0
any tips would be appreciated
Use a dictionary that has the a values as its keys so you don't have to loop through the result list for a matching a.
temp_dict = {}
for item in d:
if item['a'] in temp_dict:
temp_dict[item['a']]['b'] += item['b']
else:
temp_dict[item['a']] = item.copy()
p = list(temp_dict.values())
I have a dictionary here:
dict = {'A':['1','1','1','1','1'], 'B':['2','2'], 'C':['3','3','3','3']}
What is the necessary process to get the following result?
dict = {'A':['1_01','1_02','1_03','1_04','1_05'], 'B':['2_01','2_02'], 'C':['3_01','3_02','3_03','3_04']}
I have been learning python for quite a while now but dictionary is kind of new to me.
As others mentioned, refrain from using built-in keywords as variable names, such as dict. I kept it for simplicity on your part.
This is probably the most pythonic way of doing it (one line of code):
dict = {key:[x+"_0"+str(cnt+1) for cnt,x in enumerate(value)] for key,value in dict.items()}
You could also iterate through each dictionary item, and then each list item and manually change the list names as shown below:
for key,value in dict.items():
for cnt,x in enumerate(value):
dict[key][cnt] = x+"_0"+str(cnt+1)
Also, as some others have mentioned, if you want numbers greater than 10 to save as 1_10 rather than 1_010 you can you an if/else statement inside of the list comprehension...
dict = {key:[x+"_0"+str(cnt+1) if cnt+1 < 10 else x+"_"+str(cnt+1) for cnt,x in enumerate(value)] for key,value in dict.items()}
First iterate on keys.
Then loop on keys you are getting on key like for 'A' value is ['1','1','1','1','1'] then we can change the element at ['1','1','1','1','1']
enumerate() helps you iterate on index,value then index starts with zero as per your expected output add 1 to index. As you want the trailing 0 before each count we did '%02d'% (index+1)
Like this:
dict = {'A':['1','1','1','1','1'], 'B':['2','2'], 'C':['3','3','3','3']}
for i in dict.keys(): #iterate on keys
for index,val in enumerate(dict[i]): #took value as we have key in i
element='%02d'% (index+1) #add trailing 0 we converted 1 to int 01
dict[i][index]=val+"_"+ str(element) #assign new value with converting integer to string
print(dict)
Output:
{'A': ['1_01', '1_02', '1_03', '1_04', '1_05'], 'C': ['3_01', '3_02', '3_03', '3_04'], 'B': ['2_01', '2_02']}
Use enumerate to iterate over list keeping track of index:
d = {'A':['1','1','1','1','1'], 'B':['2','2'], 'C':['3','3','3','3']}
newd = {}
for k, v in d.items():
newd[k] = [f'{x}_0{i}' for i, x in enumerate(v, 1)]
print(newd)
Also a dictionary-comprehension:
d = {k: [f'{x}_0{i}' for i, x in enumerate(v, 1)] for k, v in d.items()}
Note: Don't name your dictionary as dict because it shadows the built-in.
d= {'A':['1','1','1','1','1'], 'B':['2','2'], 'C':['3','3','3','3']}
{x:[j + '_'+ '{:02}'.format(i+1) for i,j in enumerate(y)] for x,y in d.items()}
adict = {'A':['1','1','1','1','1'], 'B':['2','2'], 'C':['3','3','3','3'], 'D': '23454'}
newdict = {}
for i,v in adict.items():
if isinstance(v, list):
count = 0
for e in v:
count += 1
e += '_0' + str(count)
newdict[i] = newdict.get(i, [e]) + [e]
else:
newdict[i] = newdict.get(i, v)
print (newdict)
#{'A': ['1_01', '1_01', '1_02', '1_03', '1_04', '1_05'], 'B': ['2_01', '2_01', '2_02'], 'C': ['3_01', '3_01', '3_02', '3_03', '3_04'], 'D': '23454'}
This solution will check for whether your value in the dictionary is a list before assigning an index to it
You can use a dictcomp:
from itertools import starmap
d = {
'A': ['1', '1', '1', '1', '1'],
'B': ['2', '2'],
'C': ['3', '3', '3', '3']
}
f = lambda x, y: '%s_%02d' % (y, x)
print({k: list(starmap(f, enumerate(v, 1))) for k, v in d.items()})
# {'A': ['1_01', '1_02', '1_03', '1_04', '1_05'], 'B': ['2_01', '2_02'], 'C': ['3_01', '3_02', '3_03', '3_04']}
Say I have a list:
['1,2,3', '1,2,3', '1,3,2', '2,1,3', '2,3,1']
How can I categorise the items into specific keys of a dictionary according to where a specific number is in an item? If 1 is the first number, the item should be added to the value of the first key, if it's the second, it should be added to the value of the second etc.
So if I have a dictionary with keys A, B, C:
{'A': [], 'B': [], 'C': []}
The resulting dictionary should look like:
{'A': ['1,2,3', '1,2,3', '1,3,2'], 'B': ['2,1,3'], 'C':['2,3,1']
At the moment I have the following code:
lst = ['1,2,3', '1,2,3', '1,3,2', '2,1,3', '2,3,1']
dict = {'A': [], 'B': [], 'C': []}
for item in list
item.strip(',')
if item[0] == 1:
dict['A'].append(item)
elif item[1] == 1:
dict['B'].append(item)
elif item[2] == 1:
dict['C'].append(item)
print(dict)
However, this just returns the original dictionary.
try this:
lst = ['1,2,3', '1,2,3', '1,3,2', '2,1,3', '2,3,1']
dict = {'A': [], 'B': [], 'C': []}
for item in lst:
l = item.replace(',', '')
if l[0] == '1':
dict['A'].append(item)
elif l[1] == '1':
dict['B'].append(item)
elif l[2] == '1':
dict['C'].append(item)
print(dict)
I hope this helps !
I think you meant to use item.split(',') instead of item.strip(',') which only removes any commas at the start and end of the string item. item.split(',') splits the string item into a list using a comma as the delimiter. Also, you need to save the result of the method call, none of the aforementioned method calls modifies the string.
What you probably want to do is something like:
lst = ['1,2,3', '1,2,3', '1,3,2', '2,1,3', '2,3,1']
dict = {'A': [], 'B': [], 'C': []}
for item in lst:
item_arr = item.split(',')
key = 'ABC'[item_arr.index('1')]
dict[key].append(item)
print(dict)
It's less efficient and not very readable but here is a one-liner using dictionary and list comprehensions:
lst = ['1,2,3', '1,2,3', '1,3,2', '2,1,3', '2,3,1']
keys = ['A', 'B', 'C']
dic = {key: [x for x in lst if x.split(',')[j] == '1'] for j, key in enumerate(keys)}
# {'A': ['1,2,3', '1,2,3', '1,3,2'], 'B': ['2,1,3'], 'C': ['2,3,1']}
I have a Dictionary here:
dic = {'A':1, 'B':6, 'C':42, 'D':1, 'E':12}
and a list here:
lis = ['C', 'D', 'C', 'C', 'F']
What I'm trying to do is (also a requirement of the homework) to check whether the values in the lis matches the key in dic, if so then it increment by 1 (for example there's 3 'C's in the lis then in the output of dic 'C' should be 45). If not, then we create a new item in the dic and set the value to 1.
So the example output should be look like this:
dic = {'A':1, 'B':6, 'C':45, 'D':2, 'E':12, 'F':1}
Here's what my code is:
def addToInventory(dic, lis):
for k,v in dic.items():
for i in lis:
if i == k:
dic[k] += 1
else:
dic[i] = 1
return dic
and execute by this code:
dic = addToInventory(dic,lis)
It compiles without error but the output is strange, it added the missing F into the dic but didn't update the values correctly.
dic = {'A':1, 'B':6, 'C':1, 'D':1, 'E':12, 'F':1}
What am I missing here?
There's no need to iterate over a dictionary when it supports random lookup. You can use if x in dict to do this. Furthermore, you'd need your return statement outside the loop.
Try, instead:
def addToInventory(dic, lis):
for i in lis:
if i in dic:
dic[i] += 1
else:
dic[i] = 1
return dic
out = addToInventory(dic, lis)
print(out)
{'A': 1, 'B': 6, 'C': 45, 'D': 2, 'E': 12, 'F': 1}
As Harvey suggested, you can shorten the function a little by making use of dict.get.
def addToInventory(dic, lis):
for i in lis:
dic[i] = dic.get(i, 0) + 1
return dic
The dic.get function takes two parameters - the key, and a default value to be passed if the value associated with that key does not already exist.
If your professor allows the use of libraries, you can use the collections.Counter data structure, it's meant precisely for keeping counts.
from collections import Counter
c = Counter(dic)
for i in lis:
c[i] += 1
print(dict(c))
{'A': 1, 'B': 6, 'C': 45, 'D': 2, 'E': 12, 'F': 1}