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Comparing values in a variable to a dictionary in Python


By : Graeme Hendry
Date : August 01 2020, 05:00 PM
wish help you to fix your issue I have a dictionary in Python that has the name of the car, the colour and the carcount to how many is currently in stock (below is a example of a few):
code :
stock = ["Ferrari", "Lambo", "Ferrari"]

for carname in cardict.keys():
   cardict[carname]['Carcount'] = stock.count(carname)
stock = "Ferrari"
cardict[stock]['Carcount'] += 1


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Build a dictionary of values in Python based on comparing a list of dates to dates in a dictionary


By : Tony Romeo
Date : March 29 2020, 07:55 AM
may help you . Not fully sure if I get what you want but this keeps track of all placeholders and adds the second last value of total count using placeholder[-2] appends the previous value.
If you don't want the value to change until another date matches you can use a counter to keep track and use something like placeholder[-count]
code :
sql_info = {}
placeholder = []
for i,j in zip(temp_data,temp_dates):
    placeholder.append(i['total_count'])
    if i['m_date'] in temp_dates:
        sql_info[j] = i['total_count']
    else:
        sql_info[j] = placeholder[-2]
sql_info = {}
placeholder = []
count = 1
for i,j in zip(temp_data,temp_dates):
    dd = j.strftime('%m-%d-%Y')
    placeholder.append(i['total_count'])
    if i['m_date'] in temp_dates:
        sql_info[dd] = i['total_count']
    else:
        count += 1
        sql_info[dd] = placeholder[-count]
print sql_info

comparing python dictionary values


By : Borno
Date : March 29 2020, 07:55 AM
will help you There are a couple ways to achieve what you're trying to do here. I'm assuming based on the example you gave me that there are always only two words, and the lists are always ordered ordered.
No matter what the method, you'll want to iterate over the documents (The dictionary). Iterating over dictionaries is simple in Python; you can see an example here. After that, the steps change
code :
for documentNumber in docdictionary:
    for word1location in docdictionary[documentNumber][0]:
        for word2location in docdictionary[documentNumber][1]:
            if abs(word1location - word2location) == 1:
                return documentNumber
for documentNumber in docdictionary:
    list1pos = 0
    list2pos = 0
    while True:
        difference = docdictionary[documentNumber][0][list1pos] - docdictionary[documentNumber][1][list2pos]
        if abs(difference) == 1:
            return documentNumber
        if difference < 0: #Page location 2 is greater
            list1pos++
            if list1pos == len(docdictionary[documentNumber][0]): #We were at the end of list 1, there will be no more matches
                break
        else: #Page location 1 is greater
            list2pos++
            if list2pos == len(docdictionary[documentNumber][1]): #We were at the end of list 2, there will be no more matches
                break
return None

Python - comparing values in the same dictionary


By : Hannah Flor
Date : March 29 2020, 07:55 AM
will be helpful for those in need Not as elegant as using Counter, but does remove duplicates without the use of modules:
code :
d = {'Trump': ['MAGA', 'FollowTheMoney'],
    'Clinton': ['dems', 'Clinton'],
    'Stein': ['FollowTheMoney', 'Atlanta']}

dupvals = [item for sublist in d.values() for item in sublist] # get all values from all keys into a list
dups = [] # list to hold duplicates

for i in dupvals:
    if dupvals.count(i) > 1:
        dups.append(i)

dupvals = set(dups) # keep only one item for each duplicated item

new_d = {}

for key,values in d.items():
    for value in values:
        if not value in dupvals:
            new_d.setdefault(key, []).append(value)

print new_d # {'Clinton': ['dems', 'Clinton'], 'Trump': ['MAGA'], 'Stein': ['Atlanta']}

Comparing Values in a Single Dictionary - Python


By : timeline
Date : March 29 2020, 07:55 AM
fixed the issue. Will look into that further You could do the following. Collect all keys for each in dict (or defaultdict) of lists:
code :
from collections import defaultdict

temp = {"1": "hello", "2": "goodbye", "3": "hello", "4": "goodbye", "5": "hi"}
d = defaultdict(list)
for k, v in temp.items():
    d[v].append(k)

for k in d:
    print ' = '.join(repr(v) for v in d[k])  # repr only necessary to display quotes
    # print ' = '.join(d[k])
'5'
'1' = '3'
'2' = '4'

Comparing values in one dictionary with values from another dictionary in Python


By : FranceV
Date : March 29 2020, 07:55 AM
I wish did fix the issue. You've got the right idea. Iterate the items in both dicts and compare
code :
>>> dict1 = {'Canada' : 2.5, 'UK' : 3.7, 'USA' : 9.0}
>>> dict2 = {'a' : 7, 'b' : 2}
>>> 
>>> 
>>> for kd2, vd2 in dict2.items():
...     for kd1, vd1 in dict1.items():
...         if vd2 > vd1:
...             print(kd2, "is larger than", kd1)
... 
a is larger than Canada
a is larger than UK
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