给定一个整数列表，我想找到哪个数字最接近我输入的数字:
>>> myList = [4, 1, 88, 44, 3]
>>> myNumber = 5
>>> takeClosest(myList, myNumber)
...
4
有什么快速的方法吗?
当前回答
扩展Gustavo Lima的回答。不用创建一个全新的列表也可以完成同样的事情。随着FOR循环的进行，列表中的值可以替换为差值。
def f_ClosestVal(v_List, v_Number):
"""Takes an unsorted LIST of INTs and RETURNS INDEX of value closest to an INT"""
for _index, i in enumerate(v_List):
v_List[_index] = abs(v_Number - i)
return v_List.index(min(v_List))
myList = [1, 88, 44, 4, 4, -2, 3]
v_Num = 5
print(f_ClosestVal(myList, v_Num)) ## Gives "3," the index of the first "4" in the list.
其他回答
def takeClosest(myList, myNumber):
newlst = []
for i in myList:
newlst.append(i - myNumber)
lstt = [abs(ele) for ele in newlst]
print(myList[lstt.index(min(lstt))])
myList = [4, 1, 88, 44, 3]
myNumber = 5
takeClosest(myList,myNumber)
def closest(list, Number):
aux = []
for valor in list:
aux.append(abs(Number-valor))
return aux.index(min(aux))
这段代码将为您提供列表中与number最接近的数字的索引。
KennyTM给出的解决方案是最好的，但在您不能使用它的情况下(如brython)，这个函数将完成工作
>>> takeClosest = lambda num,collection:min(collection,key=lambda x:abs(x-num))
>>> takeClosest(5,[4,1,88,44,3])
4
lambda是一种书写“匿名”函数(没有名字的函数)的特殊方式。你可以给它分配任何你想要的名字，因为lambda是一个表达式。
以上内容的“长”写法是:
def takeClosest(num,collection):
return min(collection,key=lambda x:abs(x-num))
遍历列表并将当前最接近的数字与abs(currentNumber - myNumber)进行比较:
def takeClosest(myList, myNumber):
closest = myList[0]
for i in range(1, len(myList)):
if abs(i - myNumber) < closest:
closest = i
return closest
扩展Gustavo Lima的回答。不用创建一个全新的列表也可以完成同样的事情。随着FOR循环的进行，列表中的值可以替换为差值。
def f_ClosestVal(v_List, v_Number):
"""Takes an unsorted LIST of INTs and RETURNS INDEX of value closest to an INT"""
for _index, i in enumerate(v_List):
v_List[_index] = abs(v_Number - i)
return v_List.index(min(v_List))
myList = [1, 88, 44, 4, 4, -2, 3]
v_Num = 5
print(f_ClosestVal(myList, v_Num)) ## Gives "3," the index of the first "4" in the list.
