Question

嗨，大家好我正在研究我的assigemnt十个绿色瓶子，但每一个罚款，直到最后一节需要有“一个绿色的瓶子”而不是“一个绿色的瓶子”，我也将如何使用.capitalize（）

def main():
    numbers = [
    'no',
    'One',
    'Two',
    'Three',
    'Four',
    'Five',
    'Six',
    'Seven',
    'Eight',
    'Nine',
    'Ten', ]
text_one = 'green bottles\nHanging on the wall'
text_two = "\nAnd if one green bottle\nShould accidentally fall\nThere'll be"
text_three =' \n'
# Each iteration of this loop prints one verse of the song

for i in range(10, 0, -1):         
    file=open('ten.txt', 'a')
    file.write(numbers[i]+' ')
    file.write(text_one+'\n')
    file.write(numbers[i]+' ')
    file.write(text_one+' ')
    file.write(text_two+' ')
    file.write(numbers[i-1]+' ')
    file.write(text_one+'\n')
    file.write('\n')
    file.close()

  if __name__ == '__main__':
  main()

Answer 1

这会让你更接近：

numbers = ['Ten', 'Nine', 'Eight', 'Seven', 'Six', 'Five', 'Four', 'Three', 'Two', 'One', 'no']
# variable to add s when needed
plural = "s"

# loop to all but last elements starting at Ten
for i, n in enumerate(numbers[:-1]):
    # if n is One we we don't need s
    if i > 8:
        plural = ""
    # get next element for second part of string
    nxt = numbers[i+1]
    text_one = '{w} green bottle{pl}\nHanging on the wall'.format(w=n, pl=plural)
    # if the next element is One we also need don't need s for text_two
    if nxt == "One":
        plural  = ""
    text_two = "And if one green bottle\nShould accidentally fall\nThere'll be {nxt} green bottle{pl}"\
        .format(nxt=nxt.lower(),pl=plural)
    print(text_one+text_two+' hanging on the wall\n')

您需要根据列表中的位置向s添加bottle。同样重要的是下一个元素是什么，当n为2时，我们需要bottles和bottle。

要写入文件，只需使用write：

替换print

numbers = ['Ten', 'Nine', 'Eight', 'Seven', 'Six', 'Five', 'Four', 'Three', 'Two', 'One', 'no']
plural = "s"

with open("out.txt","w") as f:
    for i, n in enumerate(numbers[:-1]):
        if i > 8:
            plural = ""
        nxt = numbers[i+1]
        text_one = '{w} green bottle{pl}\nHanging on the wall'.format(w=n, pl=plural)
        if nxt == "One":
            plural = ""
        text_two = "And if one green bottle\nShould accidentally fall\nThere'll be {nxt} green bottle{pl}"\
        .format(nxt=nxt.lower(),pl=plural)
        f.write(text_one+text_two+' hanging on the wall\n\n')

Answer 2

这是另一种令人讨厌的不整洁的做法。对现有代码进行最小的更改。

基本上它允许复数s在字符串中是可选的。

text_one = 'green bottle%s\nHanging on the wall'

%s充当s的占位符，必须在以后知道使用

之类的迭代时填写

f.write(text1 % 's')

要包含s或

f.write(text1 % '')

不包含s。

完整代码：

def main():
    numbers = [
    'no',
    'One',
    'Two',
    'Three',
    'Four',
    'Five',
    'Six',
    'Seven',
    'Eight',
    'Nine',
    'Ten']
    text_one = 'green bottle%s\nHanging on the wall'
    text_two = "\nAnd if one green bottle\nShould accidentally fall\nThere'll be"
    text_three =' \n'
    # Each iteration of this loop prints one verse of the song

    with open('ten.txt', 'w') as f:
        for i in range(10, 0, -1):         

            f.write(numbers[i]+' ')
            if i == 1:
                f.write(text_one % '' +'\n')
            else:
                f.write(text_one % 's'+'\n')

            f.write(numbers[i]+' ')
            if i == 1:
                f.write(text_one % '' +'\n')
            else:
                f.write(text_one % 's'+'\n')
            f.write(text_two+' ')
            f.write(numbers[i-1]+' ')
            if (i - 1) == 1:
                f.write(text_one % '' +'\n')
            else:
                f.write(text_one % 's'+'\n')
            f.write('\n\n')

if __name__ == '__main__':
    main()

此代码不是很易维护，但它应该演示该技术。

Python 3.x十个绿色瓶子=修复复数和大写

2 个答案: