Python:如何在列表中找到小于目标的值

时间:2011-03-18 01:07:24

标签: python

例如我有一个无序的值列表[10,20,50,200,100,300,250,150]

我有这个代码,它返回下一个更大的值:

def GetNextHighTemp(self,  temp,  templist):
    target = int(temp)
    list = []
    for t in templist:
        if t != "":
            list.append(int(t))
    return str(min((abs(target - i), i) for i in list)[1])

e.g。如果temp = 55,它将返回'100'。

但我怎样才能获得较小的价值?那是如何让它返回'50'?

谢谢。

编辑 - 现在正在运作

def OnTWMatCurrentIndexChanged(self):
    self.ClearTWSelectInputs()
    material = self.cb_TW_mat.currentText()
    temp = self.txt_design_temp.text()
    if material != "":
        Eref = self.GetMaterialData(material,  "25",  "elast")
        if Eref and Eref != "":
            Eref = str(float(Eref) / 1000000000)
            self.txt_TW_Eref.setText(Eref)
        else:
            self.txt_TW_Eref.setText("194.8")
            self.ShowMsg("No temperature match found for E<sub>ref</sub> in material data file. Value of 194.8 GPa will be used.",  "blue")
    if material != "" and temp != "":
        if self.CheckTWTemp(material,  temp):
            dens = self.GetMaterialData(material,  temp,  "dens")
            self.txt_TW_dens.setText(dens)
            elast = self.GetMaterialData(material,  temp,  "elast")
            elast = str(float(elast) / 1000000000)
            self.txt_TW_Et.setText(elast)
            stress = self.GetMaterialData(material,  temp,  "stress")
            stress = str(float(stress) / 1000000)
            self.txt_TW_stress_limit.setText(stress)
        else:
            self.ShowMsg("No temperature match found for " + temp + "&#x00B0; C in material data file. Extrapolated data will be used where possible or add new material data.",  "blue")
            dens = self.GetExtrapolatedMaterialData(material,  temp,  "dens")
            self.txt_TW_dens.setText(dens)
            elast = self.GetExtrapolatedMaterialData(material,  temp,  "elast")
            elast = str(float(elast) / 1000000000)
            self.txt_TW_Et.setText(elast)
            stress = self.GetExtrapolatedMaterialData(material,  temp,  "stress")
            stress = str(float(stress) / 1000000)
            self.txt_TW_stress_limit.setText(stress)
    else:
        self.ClearTWSelectInputs()

def CheckTWTemp(self, matvar, tempvar):
    for material in self.materials:
        if material.attrib["name"] == matvar:
            temps = material.getiterator("temp")
            for temp in temps:
                if int(temp.text) == int(tempvar):
                    return True
            return False

def GetMaterialData(self, matvar, tempvar, tag):
    for material in self.materials:
        if material.attrib["name"] == matvar:
            temps = material.getiterator("temp")
            for temp in temps:
                if temp.text == tempvar:
                    value = temp.find(tag)
                    return value.text

def GetExtrapolatedMaterialData(self, matvar, tempvar, tag):
    try:
        templist = QStringList()
        for material in self.materials:
            if material.attrib["name"] == matvar:
                temps = material.getiterator("temp")
                for temp in temps:
                    templist.append(temp.text)
        templist.sort()
        target = int(tempvar)
        x1 = max(int(t) for t in templist if t != '' and int(t) < target)
        x2 = min(int(t) for t in templist if t != '' and int(t) > target)
        y1 = float(self.GetMaterialData(matvar, str(x1), tag))
        y2 = float(self.GetMaterialData(matvar, str(x2), tag))
        x = target
        y = y1 - ((y1 - y2) * (x - x1) / (x2 - x1))
        return str(y)
    except Exception, inst:
        return "0"

6 个答案:

答案 0 :(得分:18)

更好更快(代码和cpu方式)的方法是使用bisect模块进行二进制搜索,但为此您需要先对列表进行排序,这是示例用法:

import bisect

mylist = [10, 20, 50, 200, 100, 300, 250, 150]
mylist.sort()

index = bisect.bisect(mylist, 55)
print "Greater than target", mylist[index]
print "Smaller than or equal to target", mylist[index-1]

输出:

Greater than target 100
Smaller than or equal to target 50

此外,您需要检查返回的索引,如果它是0,则表示您已通过目标低于最低值

答案 1 :(得分:4)

修改:啊,我使用templist代替list - 因此造成了混乱。我并不是说它是一个单行的功能;你仍然需要进行转换。 (当然,正如Mike DeSimone正确地指出的那样,使用list作为变量名是一个糟糕的主意!!所以我有充分的理由感到困惑。:)

为了更明确一点,这里是一个稍微简化的函数版本(修复为正确测试空列表):

def GetNextHighTemp(self, temp, templist):
    templist = (int(t) for t in templist if t != '')
    templist = [t for t in templist if t < int(temp)]
    if templist: return max(templist)
    else: return None                   # or raise an error

感谢Mike建议在空列表的情况下返回None - 我喜欢这样。

你可以更加简短地缩短它:

def GetNextHighTemp(self, temp, templist):
    try: return str(max(int(t) for t in templist if t != '' and int(t) < int(temp)))
    except ValueError: return None      # or raise a different error

答案 2 :(得分:2)

nextHighest = lambda seq,x: min([(i-x,i) for i in seq if x<=i] or [(0,None)])[1]
nextLowest  = lambda seq,x: min([(x-i,i) for i in seq if x>=i] or [(0,None)])[1]

以下是它的工作原理:查看nextHighest,min的参数是列表推导,它计算列表中每个值与输入x之间的差异,但仅适用于那些值&gt; = x。由于您需要实际值,因此我们需要列表元素以包含值与实际值的差异。元组按值从左到右进行比较,因此序列中每个值i的元组变为(i-x,i) - 最小元组将具有[1]'中的实际值元素。

如果输入x值超出seq中的值范围(或者如果seq只是空),那么列表推导将给我们一个空列表,这将在min中引发一个ValueError。如果发生这种情况,我们会在or [(0,None)]的参数中添加min字词。如果列表推导为空,则它将评估为False,在这种情况下,min将查看包含单个元组(0,None)的序列。在这种情况下,[1]'元素为None,表示seq中没有高于x的元素。

以下是一些测试用例:

>>> t = [10, 20, 50, 200, 100, 300, 250, 150]
>>> print nextHighest(t,55)
100
>>> print nextLowest(t,55)
50
>>> print nextHighest([],55)
None
>>> print nextLowest([],55)
None
>>> print nextHighest(t,550)
None

答案 3 :(得分:1)

让无序列表为myList

answer = max(x for x in myList if x < temp)

答案 4 :(得分:0)

如果我理解正确,你想要的最大价值低于你的目标;例如在你的例子中,如果你的目标是55,你想要50,但如果你的目标是35,你想要20。以下函数应该这样做:

def get_closest_less(lst, target):
    lst.sort()
    ret_val = None
    previous = lst[0]
    if (previous <= target):
        for ndx in xrange(1, len(lst) - 1):
            if lst[ndx] > target:
                ret_val = previous
                break
            else:
                previous = lst[ndx]
    return str(ret_val)

如果您需要单步执行这些值,可以使用generator连续获取值:

def next_lesser(l, target):
    for n in l:
        if n < target:
            yield str(n)

这些都可以在一个简单的程序中正常工作。

答案 5 :(得分:0)

a=[4,3,8,2,5]
temp=4
def getSmaller(temp,alist):
    alist.sort()
    for i in range(len(alist)):
        if(i>0 and alist[i]==temp):
            print alist[i-1]
        elif(i==0 and alist[i]==temp):
            print alist[i]
getSmaller(temp,a)