Question

所以我有两个变量mean和service由以下值组成：

mean=[5.76,6.5,7,5,4.5,6,9,2.5,3,5.5]
limit=[90,80,95,96,98,85,82,75,90,91]

现在，我在变量mean上应用了poisson函数：

from scipy.stats import poisson
for m in mean:
    r_v=poisson(m)
    r_v1.append(r_v)

接下来我需要做的是获取所有均值的pmfs的总和（从0开始的出现次数）并将它们与limit中的相应值进行比较。我需要获取pmfs之和大于或等于limit中相应值的出现次数。例如，对于平均值= 5.76，（pmf（0）+ pmf（1）... + pmf（9））* 100 = 93.2，其在9个出现的极限变量中超过90。我运行以下代码似乎工作正常：

for r,s in zip(r_v1, limit):
        l=[]
        z=0
        while True:
            y=r.pmf(z)
            l.append(y)
            z+=1
            if round(sum(l),2)*100>=s:
                break
Store=%z

问题是，当我尝试存储round(sum(l),2)*100>=s的最后一次出现的z时，我收到以下错误：

ERROR:root:Line magic function `%z` not found.

有人可以帮我纠正代码中的错误吗？非常感谢。

Answer 1

%是在IPython的交互式shell中调用魔术行函数的前缀，并且没有z行魔术函数，因此您会收到此错误。

在IPython中输入%quickref，您可以了解可用的魔术函数或readthedocs。

你的问题是你在一步中尝试了太多。尝试为一个输入获得正确的结果，然后在另一个步骤中累积：

def calc(mean_, limit_):

    def inner(m, s):
        r = poisson(m)
        l = []
        z = 0
        while True:
            y = r.pmf(z)
            l.append(y)
            z += 1
            if round(sum(l), 2) * 100 >= s:
                return z, l

    return [inner(*args) for args in zip(mean_, limit_)]

致电calc(mean, limit)给出：


[(10,
  [0.0031511115984444414,
   0.018150402807039979,
   0.052273160084275148,
   0.10036446736180832,
   0.14452483300100394,
   0.16649260761715656,
   0.15983290331247035,
   0.13151964615426115,
   0.094694145231068075,
   0.060604252947883679]),
 (10,
  [0.0015034391929775724,
   0.009772354754354215,
   0.031760152951651209,
   0.068813664728577653,
   0.11182220518393866,
   0.14536886673912017,
   0.1574829389673803,
   0.14623415761256733,
   0.11881525306021086,
   0.08581101609904139]),
 (12,
  [0.00091188196555451624,
   0.0063831737588816145,
   0.022341108156085643,
   0.052129252364199796,
   0.091226191637349643,
   0.1277166682922895,
   0.14900277967433773,
   0.14900277967433773,
   0.1303774322150455,
   0.10140466950059107,
   0.070983268650413558,
   0.045171170959354162]),
 (10,
  [0.006737946999085467,
   0.033689734995427337,
   0.084224337488568321,
   0.1403738958142805,
   0.17546736976785063,
   0.17546736976785068,
   0.1462228081398754,
   0.10444486295705395,
   0.065278039348158651,
   0.036265577415643714]),
 (10,
  [0.011108996538242306,
   0.049990484422090385,
   0.11247858994970336,
   0.168717884924555,
   0.18980762054012446,
   0.17082685848611215,
   0.1281201438645839,
   0.082362949627232548,
   0.046329159165318316,
   0.0231645795826592]),
 (9,
  [0.0024787521766663585,
   0.014872513059998144,
   0.044617539179994441,
   0.089235078359988937,
   0.13385261753998332,
   0.16062314104797995,
   0.16062314104798009,
   0.13767697804112569,
   0.10325773353084421]),
 (13,
  [0.00012340980408667956,
   0.0011106882367801166,
   0.0049980970655105232,
   0.014994291196531574,
   0.033737155192196056,
   0.06072687934595293,
   0.091090319018929264,
   0.1171161244529091,
   0.13175564000952278,
   0.13175564000952278,
   0.11858007600857066,
   0.097020062188830414,
   0.072765046641622894]),
 (4,
  [0.0820849986238988,
   0.20521249655974699,
   0.25651562069968376,
   0.21376301724973648]),
 (6,
  [0.049787068367863944,
   0.14936120510359185,
   0.22404180765538775,
   0.22404180765538775,
   0.16803135574154085,
   0.10081881344492458]),
 (10,
  [0.0040867714384640666,
   0.02247724291155237,
   0.06181241800676901,
   0.1133227663457432,
   0.15581880372539689,
   0.17140068409793663,
   0.15711729375644187,
   0.12344930223720431,
   0.084871395288077939,
   0.051865852676047694])]

将最后一次迭代存储在循环中（python）

1 个答案: