小编Nil*_*esh的帖子

我正在创建一个python脚本,打印出'99瓶啤酒'的整首歌曲,但是相反.我唯一无法逆转的是数字,整数,而不是字符串.

这是我的完整脚本,

def reverse(str):
   return str[::-1]

def plural(word, b):
    if b != 1:
        return word + 's'
    else:
        return word

def line(b, ending):
    print b or reverse('No more'), plural(reverse('bottle'), b), reverse(ending)

for i in range(99, 0, -1):
    line(i, "of beer on the wall")
    line(i, "of beer"
    print reverse("Take one down, pass it around")
    line(i-1, "of beer on the wall \n")

我理解我的反向函数将字符串作为参数,但是我不知道如何接受整数,或者如何在脚本中反转整数.

doctest很难控制。我遇到了这样的问题

功能

from collections import namedtuple

Match = namedtuple('Match', ['token_string', 'normalised_token',
                     'brand_name', 'brand_id',
                     'score'])


def make_match(tokens, normalised, brand, score):
"""
Examples:
>>> make_match('Jack Jones','JackJones',('Jack Jones','X023'),0.6)==Match('Jack Jones','JackJones','Jack Jones','X023',0.6)
True 
>>> make_match('Jack Jones','JackJones',('Jack Jones','X023'),0.6)==('Jack Jones','JackJones','Jack Jones','X023',0.6)
True
>>> match=make_match('Jack Jones','JackJones',('Jack Jones','X023'),0.6)
>>> match.token_string=='Jack Jones'
True
"""
return Match(token_string=tokens,
         normalised_token=normalised,
         brand_name=brand[0],
         brand_id=brand[1],
         score=score)

但是有一个错误

Failed example:
make_match('Jack Jones','JackJones',('Jack Jones','X023'),0.6)==Match('Jack Jones','JackJones','Jack Jones','X023',0.6)
Expected:
    True 
Got:
    True

1个项目有故障：

难道不是完全匹配了吗？非常感谢利用率4中的1。make_match 测试失败 1个失败。

使用Python NumPy计算矩阵列的总和:

import numpy
from StringIO import StringIO

fileName = 'test2.csv'
myFile = open(fileName,'r')
print "Reading data from '%s' ..." % fileName
data = myFile.read() 
myFile.close()

data = numpy.genfromtxt(StringIO(data), delimiter=',', usecols=(0,1,2))
print "Calculating ..."

print data[:,2]
sumA1 = data[:,2].sum
print "shape =", data.shape
print "sumA1 =", str(sumA1)
print "ok"

然后内容test2.csv:

12,13,14,17
1,2,3,4
12,13,14,17
1,2,3,4
12,13,14,17
1,2,3,4
12,13,14,17
1,2,3,4
12,13,14,17

我得到了这个输出

Reading data from 'test2.csv' ...
Calculating ...
[ 14.   3.  14.   3.  14.   3.  14.   3.  14.]
shape …

我有声明类的功能

>>> def a():
...     class A:
...         print "a"
...

当我调用函数时,为什么要打印a？

>>> a()
a

我想在我的剧本依赖项中添加一个角色，但要基于条件。

- name: Get all install pyenv versions
  command: '{{ pyenv_root }}bin/pyenv versions'
  register: available_versions
  tags:
      - get_pyenv_versions
  environment:
      PYENV_ROOT: "{{ pyenv_root }}"

dependencies:
    - { role: pyenv, python_versions: ["{{ mypython_version }}"], when: "mypython_version not in available_versions.stdout" }

我想做的是，我想检查所有可用pyenv版本，如果mypython_version不可用，那么只有我想调用pyenv角色，否则我不想调用它。

它给了我语法错误

ERROR! Syntax Error while loading YAML.


The error appears to have been in '/root/ansible_playbooks/roles/mydeployment/meta/main.yaml': line 9, column 1, but may
be elsewhere in the file depending on the exact syntax problem.

The offending line …

我刚刚开始学习Python,并试图调用我在另一个函数中定义的一个函数.解释器返回错误:NameError:未定义全局名称'isPrime'

这是我的代码的样子:

def split (value):
    x = 2
    halfOne, halfTwo = 0, 0
    if isPrime(value) == True:
        print (value)
        return
    else:
        while x < value:
            if value % x == 0:
                halfOne = value / x
                halfTwo = x
                split(halfOne)
                split(halfTwo)
            x += 1
def is_prime(value):
# some code

感谢您的帮助!