공부하는 개발 블로그

리스트의 원소 숫자 세기 : counting number of elements in the list

ex. ["a", "b", "c" ,"a", "c", "a" ] -> [("a", 3), ("b",1), ("c", 2)]

def count(L : list) = L.groupby(identity).mapValues(_.size)

https://leetcode.com/problems/find-all-numbers-disappeared-in-an-array/submissions/

class Solution: 
    def findDisappearedNumbers(self, nums: List[int]) -> List[int]: 
        for i in range(len(nums)): 
            c = 0 
            while nums[i] != nums[nums[i]-1] and nums[i] != i+1 : 
                if c > 4: break 
                print(nums, nums[i], nums[nums[i]-1])
                # right
                nums[i], nums[nums[i]-1] = nums[nums[i]-1], nums[i] 
                # wrong
                #nums[nums[i]-1], nums[i] = nums[i], nums[nums[i]-1] 
                c+=1 

dict.pop(key, None)

D_index = {k: v for k, v in enumerate(D)}

Make pairs

returns iterator

x = ['a', 'b', 'c']
y = [1, 2]

zipped = list(zip(x, y))
# [('a', 1), ('b', 2)]

x2, y2 = zip(*zip(x, y))
# x2 = ['a', 'b']
# y2 = [1, 2]