Compare two lists which are values in a dict and return which has the min val in the list Python

Question

I have a dict like this

b = {'2': ['10', '5', '4'], '4': ['1', '9', '2'], '3': ['90', '87', '77'], '1': ['30']}

I need to compare each value in the list to others and return only the least value in the dict

I have tried

for k,v in b.items():
  for r in range(len(v)):
    print(min(v[r] + v[r]))

It is giving me a weird output! This is the output obtained from that code.

0
5
4
1
9
2
0
7
7
0
0
0
0

I need the key and value which has the least value in the entire dict output like this d = {4:[1]}

Answer 1

Ugly one-liner:

b =  {'2': ['10', '5', '4'], '4': ['1', '9', '2'], '3': ['90', '87', '77'], '1': ['30']}
result = dict([min(((int(k), [min(map(int, v))]) for k, v in b.items()), key=lambda t: t[1])])
print(result)

Output:

{4: [1]}

Breakdown:

b =  {'2': ['10', '5', '4'], '4': ['1', '9', '2'], '3': ['90', '87', '77'], '1': ['30']}

# Generator of each key with its minimal element
# (here the generator would produce the list [(2, [4]), (4, [1]), (3, [77]), (1, [30])])
key_min = ((int(k), [min(map(int, v))]) for k, v in b.items())
# Pick tuple with minimal value
# (here the tuple (4, [1]) from the previous generator)
min_entry = min(key_min, key=lambda t: t[1])
# Make into dict
# (here {4: [1]}; first element of the tuple is the key and second element the value)
result = dict([min_entry])
print(result)

Answer 2

您可以通过dict理解来做到这一点

 {int(key): [min( int(value) for value in value_list)] for key, value_list in b.items()}

Answer 3

If you want a straightforward answer without any confusion

min_list = {}
for k,v in b.items():
    min_value = min(v)
    min_list[min_value] = k

print({ min_list[min(min_list)]:min(min_list)})

Answer 4

You want the minimum of minimums, or:

min({k: min(b[k], key=int) for k in b}.items(), key=lambda x: x[1])

This returns the tuple ('4', '1') .

Answer 5

First, your list is numbers as text. I did not correct that. If you can fix that then you can take off the in() in this code.

 for k, v in b.items():
    x = int(min(b[k]))
    try:
        lowVal
    except:
        lowVal = x
        lowKey = k
    else:
        if x < lowVal:
            lowKey = k
            lowVal = x
 print('{0}: {1}'.format(lowKey, lowVal))

1. Step through each item in the dict
2. find the lowest value and make it an int() and set to x for convenience
3. try to see if this is our first time through, if it is set the key to lowKey and the lowest value in the list to lowVal
4. Otherwise if lowVal already exists see if the current lowest value in the list is lower than the previous lowest. If it is then set lowKey and lowVal to the current loops values
5. Print
6. ????
7. Profit

Edit: a word

Answer 6

What if you have multiple key value pairs with same minimum value?

This solution works fine for that as well.

result={k:min(map(int,v)) for k,v in b.items()}
minVal=min(result.values())
result={k:[minVal] for k in result.iterkeys()  if result[k] == minVal}
print(result)
{'4': [1]}

for ex :

b = {'2': ['10', '5', '4'], '4': ['1', '9', '2'], '3': ['90', '1', '77'], '1': ['30']}

Output will be :

{'3': [1], '4': [1]}