将两个由 print 语句分隔的 for 循环连接成一个

Question

在下面的代码中，一切都按预期工作。

它获取以 0 结尾的 4 个字符长的用户输入。

并且简单地将元音和辅音的出现添加到字典中。

input =""   #get input from user

while 4 < len(input) or 4 > len(input) or input[-1] != "0": #string has to be 4 char long and end in 0
    input = raw_input("insert code:")

occurrences = {"a":0,"e":0,"i":0,"o":0,"u":0,"consonants":0}    #store the vouel count



for x in input:
    if x in occurrences.keys():
        occurrences[x] += 1  #count cowels
    elif x != "0":
        occurrences["consonants"] += 1   #count consonants


for x in occurrences:
    if occurrences[x] > 0 and x != "consonants":
        print x + ",",

print "were inserted",


for x in occurrences:
    if occurrences[x] > 0 and x != "consonants":
        print str(occurrences[x]) + ",",

print "times respectively"



if occurrences["consonants"] == 1:
    print "there was %d consonant"%occurrences["consonants"]
else:
    print "there was %d consonants"%occurrences["consonants"]

对于输入“aef0”，程序将打印：

e, a, 分别插入 1, 1, 次
分别有1个辅音

我的问题是关于这一特定行的。

我知道一定有更好的方法：

for x in ocurrances:
    if ocurrances[x] > 0 and x != "consonants":
        print x + ",",

print "were inserted",


for x in ocurrances:
    if ocurrances[x] > 0 and x != "consonants":
        print str(ocurrances[x]) + ",",

print "times respectively"

只是感觉很草率。

我不喜欢的是，我调用了两次相同的 for 循环，我觉得这可能只是以一种更优雅的方式进行的一个举动，但我没有找到这样做的方法。

我想要实现的伪代码（或其他）如下。

loop the dictionary
print all key with values >= 1
print "were inserted" only once
print all the respective vales.
print "times respectively"

正如我所说，我想要相同的输出，但以更优雅的方式表达，我假设优雅意味着只有一个 for 循环，但欢迎任何其他（更优雅）的选项！

我想过做类似this的事情，但显然行不通。 （别担心，这完全是错误的，但该方法显示了我的目标）

提前致谢！

Answer 1

编写代码的另一种方法可能是这样的：

print ", ".join(k for k, v in occurrences.items() if k != "consonants" and v > 0),
print "were inserted"
print ", ".join(str(v) for k, v in occurrences.items() if k != "consonants" and v > 0),
print "times respectively"

您可以通过分解搜索来缩短它：

a = [(k, str(v)) for k, v in occurrences.items() if k != "consonants" and v > 0]
print ", ".join(x[0] for x in a), "were inserted",
print ", ".join(x[1] for x in a), "times respectively"

Answer 2

你在优雅和其他更重要的事情上还有一些问题。首先，您的用户可能会反对必须键入 0，因为您没有用它做任何有意义的事情。事实上，您需要专门编写代码来忽略它！在字典中保留辅音的计数是相当不雅的。您不检查用户是否输入了所有字母。你不处理大写。

下面是一些解决这些问题的代码以及一些冗长的内容：

input = ""   # get input from user
while len(input) != 3 or not input.isalpha():
    input = raw_input("insert code:").lower()
ocurrances = {"a":0, "e":0, "i":0, "o":0, "u":0}
nconsonants = 0
for x in input:
    if x in ocurrances:
        ocurrances[x] += 1  #count vowels
    else:
        nconsonants += 1   #count consonants
for x in ocurrances:
    if ocurrances[x]:
        print x + ",",
print "were inserted",
for x in ocurrances:
    if ocurrances[x]:
        print str(ocurrances[x]) + ",",
print "times respectively"
if nconsonants == 1:
    print "there was 1 consonant"
else:
    print "there were %d consonants" % nconsonants

您可能希望将“ocurrences”更改为“occurrences”。顺便说一句，如果元音的数量少于 2 个，输出就不是很漂亮。