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
如果我可以补充@Lauritz的回答
为了不出现运行错误 不要忘记在bisect_left行之前添加一个条件:
if (myNumber > myList[-1] or myNumber < myList[0]):
return False
所以完整的代码看起来像这样:
from bisect import bisect_left
def takeClosest(myList, myNumber):
"""
Assumes myList is sorted. Returns closest value to myNumber.
If two numbers are equally close, return the smallest number.
If number is outside of min or max return False
"""
if (myNumber > myList[-1] or myNumber < myList[0]):
return False
pos = bisect_left(myList, myNumber)
if pos == 0:
return myList[0]
if pos == len(myList):
return myList[-1]
before = myList[pos - 1]
after = myList[pos]
if after - myNumber < myNumber - before:
return after
else:
return before
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
>>> 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))
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
def find_nearest(array, value):
array = np.asarray(array)
idx = (np.abs(array - value)).argmin()
return array[idx]
通过使用
price_near_to=find_nearest(df['Close'], df['Close'][-2])
