How to get the first value in a python dictionary

Question

I have a dictionary like this:

myDict = {  
    \'BigMeadow2_U4\': (1609.32, 22076.38, 3.98),  
    \'MooseRun\': (57813.48, 750187.72, 231.25),  
    \'Hw         


        

Answer 1

myList = []  
for k,v in myDict.items()  
    myList.append(v[0])

Answer 2

Try this way:

my_list = [elem[0] for elem in your_dict.values()]

Offtop: I think you shouldn't use camelcase, it isn't python way

UPD: inspectorG4dget notes, that result won't be same. It's right. You should use collections.OrderedDict to implement this correctly.

from collections import OrderedDict
my_dict = OrderedDict({'BigMeadow2_U4': (1609.32, 22076.38, 3.98), 'MooseRun': (57813.48, 750187.72, 231.25), 'Hwy14_2': (991.31, 21536.80, 6.47) })

Answer 3

one lines...

myList = [myDict [i][0] for i in sorted(myDict.keys()) ]

the result:

>>> print myList 
[1609.32, 991.31, 57813.48]

Answer 4

If you want ordered dictionary, use:

from collections import OrderedDict
ordered = OrderedDict(
    ('BigMeadow2_U4', (1609.32, 22076.38, 3.98)),  
    ('MooseRun', (57813.48, 750187.72, 231.25)),  
    ('Hwy14_2', (991.31, 21536.80, 6.47)) 
)
first_values = [v[0] for v in ordered.values()]

The output order will be exactly as your input order.

Answer 5

for getting first value

print(getDictKeyandValue(d,1))

Answer 6

Try this.Type the dictionary and the position you want to print in the function

d = {'Apple': 1, 'Banana': 9, 'Carrot': 6, 'Baboon': 3, 'Duck': 8, 'Baby': 2}
print(d)
def getDictKeyandValue(dict,n): 
    c=0
    mylist=[]
    for i,j in d.items():
        c+=1
        if c==n:

            mylist=[i,j]
            break
    return mylist   

print(getDictKeyandValue(d,2))