如何比较 python 中列表/数组/字典中的对应元素

Question

Suppose you have two lists (or any type of grouping, it doesn't matter which one) containing variables that represent milk, eggs, and flour. For instance:

have(milk, eggs, flour)

and

need(milk, eggs, flour)

How might you go about determining whether each element >=, ==, or <= its counterpart in the other list so that you can return some indication as to whether there is or is not enough of each ingredient to make a proverbial cake or if there is enough to make more than one?

I'd really prefer not to write War and Peace for the sake of 3 comparisons. Any help is appreciated.

Answer 1

You could use dictionaries. For instance:

have = {"milk": 2, "eggs": 3, "flour": 0.5}
need = {"milk": 1, "eggs": 5, "flour": 2.5}
ingredients = {i:"Yes" if have[i] >= need[i] else "No" for i in have.keys()}

Output:

print(ingredients)
{'milk': 'Yes', 'eggs': 'No', 'flour': 'No'}

If you want a function that tells you how many cakes you can do with the ingredients you have, you can use the following:

def how_many(need, have):
    results = {i:have[i]//need[i] for i in have.keys()}
    return min(results.values())

Answer 2

If the a list of quantities that you need to compare, you can use a one-liner list comprehension (can only compare ==, >, < else it they will be overlapping operations if you use >= and <= and ==) -

milk_have = 10
eggs_have = 20
flour_have = 30

milk_need = 10
eggs_need = 25
flour_need = 3

have = [milk_have, eggs_have, flour_have]
need = [milk_need, eggs_need, flour_need]

['==' if i[0]==i[1] else '>' if i[0]>i[1] else '<' for i in zip(have, need)]

['==', '<', '>']

Answer 3

Asssuming input as the following

milk=200
eggs=10
flour=1000

milk_reqd=100
eggs_reqd=5
flour_reqd=2000

have=[milk, eggs, flour]
need=[milk_reqd, eggs_reqd, flour_reqd]

Solution

import numpy as np
have=np.array(have)
need=np.array(need)

Now you can perform all operations like

need>have

Or

need<=have

Or

need-have

To get the number of cakes that can be made

n_cakes=int(min(have/need))

Answer 4

have = ('milk', 'eggs', 'flour',"k")
need = ('milk', 'eggs', 'flour',"l")
incredients = {}
for i in range(len(have)):
   count = 0 
   if have[i] == need[i]:
      count +=1
   incredients[have[i]] = count

output: {'eggs': 1, 'flour': 1, 'k': 0, 'milk': 1}