Question

我已经完成了任务，但是以最基本的形式寻找帮助缩短它，所以它可以应用于任何一个字，而不仅仅是一个八字母，这是我到目前为止所做的事情（有点长）：

alpha = map(chr, range(97, 123))
word = "computer"
word_list = list(word)

one = word[0]
two = word[1]
three = word[2]
four = word[3]
five = word[4]
six = word[5]
seven = word[6]
eight = word[7]


one_index = str(alpha.index(one))
two_index = str(alpha.index(two))
three_index = str(alpha.index(three))
four_index = str(alpha.index(four))
five_index = str(alpha.index(five))
six_index = str(alpha.index(six))
seven_index = str(alpha.index(seven))
eight_index = str(alpha.index(eight))

print (one + "=" + one_index)
print (two + "=" + two_index)
print (three + "=" + three_index)
print (four + "=" + four_index)
print (five + "=" + five_index)
print (six + "=" + six_index)
print (seven + "=" + seven_index)
print (eight + "=" + eight_index)

Answer 1

您可能正在寻找的是 for-loop 。

使用for循环，您的代码可能如下所示：

word = "computer"

for letter in word:
  index = ord(letter)-97
  if (index<0) or (index>25):
    print ("'{}' is not in the lowercase alphabet.".format(letter))
  else:
    print ("{}={}".format(letter, str(index+1))) # +1 to make a=1

如果您使用

for letter in word:
  #code

对于单词中的每个字母（或单词中的单词，如果单词是列表），将执行以下代码。

开始了解有关循环的更多信息，请访问：https://en.wikibooks.org/wiki/Python_Programming/Loops

您可以在互联网上找到大量涉及此主题的资源。

Answer 2

用于循环for循环，

alpha = map(chr, range(97, 123))
word = "computer"
for l in word:
    print '{} = {}'.format(l,alpha.index(l.lower()))

<强>结果

c = 2
o = 14
m = 12
p = 15
u = 20
t = 19
e = 4
r = 17

Answer 3

从dict开始，将每个字母映射到其编号。

import string
d = dict((c, ord(c)-ord('a')) for c in string.lowercase)

然后将字符串的每个字母与相应的索引配对。

result = [(c, d[c]) for c in word]

Answer 4

感谢帮助设法使用函数和while循环以不同的方式自己解决它，而不是简短但适用于所有小写单词：

alpha = map(chr, range (97,123))
word = "computer"
count = 0
y = 0

def indexfinder (number):
     o = word[number]
     i = str(alpha.index(o))
     print (o + "=" + i)


while count < len(word):
    count = count + 1
    indexfinder (y)
    y = y+1

Python：将单词转换为字母列表，然后根据小写字母返回字母的索引

4 个答案: