TypeError:'float'类型的对象没有len()

时间:2013-11-27 15:40:15

标签: python

这是错误:对于范围内的j(len(rotlati)):

TypeError:'float'类型的对象没有len()

我看过很多其他帖子,但我还没有找到解决方案。 我有点困惑。如果您知道这里发生了什么,请发表评论。

代码是:

m = 22
rlati = numpy.zeros(m)
n = 22
rlongi = numpy.zeros(n)

v = numpy.ndarray((2,),float)

for j in range (len(lati)):
LA = lati[j]
    rlati[j] = LA - latiref

    for i in range (len(longi)):
        LO = longi[i]

    rlongi[i] = LO - longiref

    v[0] = rlati[j]
    v[1] = rlongi[i]

    vv = numpy.matrix(v)
    #transpose of vv as vv.T
    vv = vv.T

    #proper rotation
    vn = R*vv

    #define how many decimals
    vn = numpy.around(vn, decimals =2)

    # rotation of the second column (lati) and third line (longi)
        rotlati = float(vn[0])
    rotlongi = float(vn[1])

s = 22
latidef = numpy.zeros(s)
p = 22
longidef = numpy.zeros(p)

for j in range (len(rotlati)):
RLA = rotlati[j]

    latidef[j] = RLA + latiref

for i in range (len(rotlongi)):
        RLO = rotlongi[i]

    longidef[i]= RLO + longiref

    RLADEF = latidef[j]
    RLODEF = longidef[i]

    return RLADEF, RLODEF

4 个答案:

答案 0 :(得分:5)

错误正是它所说的。 rotlatifloat。您无法获取len()的{​​{1}}。查看您的代码,看起来好像您可能意味着来创建名为floatrotlati列表,并在每次迭代时附加到它们你的rotlongi循环。相反,您目前只是在每次迭代时覆盖相同的两个浮点变量。

答案 1 :(得分:1)

len参数可以是序列(字符串,元组或列表)或映射(字典)。 https://docs.python.org/2/library/functions.html#len

在调用len函数之前,您应该验证参数是否属于此类型之一。您可以调用方法isinstance()来验证它。看看如何使用它。 https://docs.python.org/2/library/functions.html#isinstance

答案 2 :(得分:0)

我得到了正确答案,试试这个:

lloyd = {
    "name": "Lloyd",
    "homework": [90.0, 97.0, 75.0, 92.0],
    "quizzes": [88.0, 40.0, 94.0],
    "tests": [75.0, 90.0]
}

alice = {

    "name": "Alice",
    "homework": [100.0, 92.0, 98.0, 100.0],
    "quizzes": [82.0, 83.0, 91.0],
    "tests": [89.0, 97.0]
}
tyler = {

    "name": "Tyler",
    "homework": [0.0, 87.0, 75.0, 22.0],
    "quizzes": [0.0, 75.0, 78.0],
    "tests": [100.0, 100.0]
}

numbers = []

def average(numbers):

    total = float(sum(numbers))
    return total / len(numbers)

答案 3 :(得分:0)

另一种情况是我在保存文件时遇到了同样的错误。

odf.to_excel(writer,sheet_name = str(x),startcol = 5,index = False)

您还可以检查工作表的名称是int还是float。将其转换为字符串。