我得到这个错误,不能解决它,也不能在互联网上找到它。
TypeError: 'numpy.float64' object cannot be interpreted as an integer
TypeError Traceback (most recent call last)
<ipython-input-10-33f2a17ec582> in <module>
20 print("Saving New CSV file")
2
我尝试了一些关于do和out的类型转换的例子。我不明白为什么下面的代码片段不能输出正确的结果。
/* int to float */
#include<stdio.h>
int main(){
int i = 37;
float f = *(float*)&i;
printf("\n %f \n",f);
return 0;
}
这将打印0.000000
/* float to short */
#include<stdio.h>
int main(){
float
我有一个试图在循环中访问的子图列表:
index=[5,3,4,1,1,3,4,2,3,4,2,2,3,3,2,4]
subgraph=[[subgraph1],[subgraph2],[subgraph3],[subgraph4],[subgraph5]]
for i in range(len(index)):
for j in range(i+1,len(index)):
if index[j]==index[i]
continue
testgraphi=copy.copy(subgraph[index[
我目前有一个numpy多维数组(类型为float)和一个numpy列数组(类型为int)。我想把它们组合成一个多维的numpy数组。
import numpy
>> dates.shape
(1251,)
>> data.shape
(1251,10)
>> test = numpy.hstack((dates, data))
ValueError: all the input arrays must have same number of dimensions
要显示数组类型不同,请执行以下操作:
>> type(dates[0])
<
代码:
import pickle
test = 3
>>> with open('test', 'wb') as file:
... pickle.dumps(test, file)
和意外报告的错误。
Traceback (most recent call last):
File "<stdin>", line 2, in <module>
TypeError: an integer is required (got type _io.BufferedWriter)
这里发生了什么事?
我正在尝试使用烧瓶-python连接mysql数据库。我创建了一个“用户”名称表,在我的数据库名“login”中有两个列“fname”'lname‘数据类型字符串
代码
from flask import Flask
from flask_mysqldb import MySQL
app=Flask(__name__)
app.config['MYSQL_HOST']='localhost'
app.config['MYSQL_PORT']='3306'
app.config['MYSQL_
我试图在一个区间中创建线性函数,而另一个区间是0。并将其添加到数组中。
所以我尝试了这段代码
import numpy as np
import sympy as sp
import matplotlib as plt # This is all the library's i need
import mpmath
x = sp.symbols('x')
n = 10
xx = (np.array([np.linspace(0, 1, n+1)]))
i = 0
N = []
N[0] = sp.Piecewise( ((xx[0, i + 1] - x) / (xx
试图绘制图时,我从matplotlib中得到了以下错误
TypeError: %d format: a number is required, not numpy.float64
这是完整的跟踪(我修改了路径名称):
Traceback (most recent call last):
File ".../plotmod.py", line 154, in _plot
fig.autofmt_xdate()
File ".../local/lib/python2.7/site-packages/matplotlib/figure.py", lin
int n ;
n= (int)( javax.swing.JOptionPane.showInputDialog(null,"enter a 3 digit no."));
为什么上面给出的是errorrequired int,found string,而下面的工作正常?
int n ;
n= Integer.parseInt( javax.swing.JOptionPane.showInputDialog(null,"enter a 3 digit no."));
我想把两个数字相乘,我知道我的数字总是正数,然后:
unsigned int mulPositiveNumbers(unsigned int a ,unsigned int b)
{
assert(a > 0);
assert(b > 0);
return (a*b);
}
现在,我使用assert告诉自己“给定的数字总是正数”。
但是当我运行的时候:
int main()
{
unsigned int res = mulPositiveNumbers(-4,3);
// more stuff goes here
}
即使我使用
编辑:这里的答案:帮助理解这个问题。然而,这个问题并不是语言不可知论。它特定于SQLite处理的浮点数的文档化行为和亲和力。有一个与另一个问题非常相似的答案。
问:我有一个相当复杂的SQLite,其中子句比较数值。我在这里读过并“思考”过数据类型文档:
对于SQLite用于确定比较子句(如=、>、<、<>等)中的数据类型的逻辑仍然感到困惑。我可以将我的示例缩小到这个测试SQL中,其结果对我来说没有什么意义。
SELECT
CAST(10 AS NUMERIC) + CAST(254.53 AS NUMERIC) = CAST(264.53 AS NUMERIC)
我正在查看具有以下if-test的第三方库:
if isinstance(xx_, numpy.ndarray) and xx_.dtype is numpy.float64 and xx_.flags.contiguous:
xx_[:] = ctypes.cast(xx_.ctypes._as_parameter_,ctypes.POINTER(ctypes.c_double))
看起来xx_.dtype is numpy.float64总是失败:
>>> xx_ = numpy.zeros(8, dtype=numpy.float64)
>>>
我正在查询一个使用Binary(20)列存储主键(UUID)值的表。为什么在WHERE子句中使用速记时,二进制数据被认为是falsey?
# This returns 0 records:
SELECT
*
FROM
my_table
WHERE
primary_uuid
当显式地声明WHERE约束时,不被认为是falsey:
# This query returns all rows in the table
SELECT
*
FROM
my_table
WHERE
primary_uuid IS NOT NULL
/* OR !primary
它显示60 +5是605,而不是65
g = document.getElementById("height1").value * 12;
alert(g); <---- This is showing 60
h = document.getElementById("height2").value;
alert(h); <---- This is showing 5
b = g+h;
alert(b); <---- This is showing 605
任何想法
我正在尝试运行下面的脚本,我认为(Fmin)有错误。我有以下错误:
TypeError:“numpy.float64”对象不可调用
非常感谢你,我试了很多,但我不能再次...Thanks
import numpy as n
from scipy import optimize
a=2
b=3
def f (ts):
c= ts
y= optimize.fmin(np.linalg.norm(a/c +b),x0=0.1)
return y
f2=n
我编写了以下小程序来打印斐波纳契序列:
static void Main(string[] args)
{
Console.Write("Please give a value for n:");
Int16 n = Int16.Parse(Console.ReadLine());
Int16 firstNo = 0;
Int16 secondNo = 1;
Console.WriteLine(firstNo);
Console.WriteLine(secondNo);
for (Int16 i = 0; i <
当输入时,它会说:
Traceback (most recent call last):
File "/Users/marco/PycharmProjects/pythonProject/goncas.py", line 4, in <module>
print("A pessoa que tem "+ idade2 +" e mais velha que a pessoa que tem "+ idade)
TypeError: cannot concatenate 'str' and 'int&
我正在尝试这个例子:
p = 10000
n = 12
r = 8
t = int(input("Enter the number of months the money will be compounded "))
a = p (1 + (r/n)) ** n,t
print (a)
。。但错误是:
TypeError: 'int' object is not callable
Python是否将p视为函数?如果是这样的话,有没有办法不导入模块就能做到呢?
谢谢!