将两个由打印语句分隔的for循环合并为一个

Question

在下面的代码中，一切都按预期工作。

它获取一个长度为4的用户输入，最后以0结尾。

然后简单地在一个字典中记录元音和辅音出现的次数。

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”，程序会打印：

e, a, 各出现了1次
而且有1个辅音

我想问的是关于这几行代码。

我知道一定有更好的方法来做：

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"

我觉得这样写有点乱。

我不喜欢的是，我调用了两次相同的循环，我觉得可以用一种更优雅的方式只用一次循环，但我找不到方法。

我想要实现的伪代码（或者其他什么）大概是这样的。

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，但用更优雅的方式表达，我假设优雅的方式应该只用一个循环，但其他任何（更优雅的）选项也欢迎！

我考虑过做类似这个的事情，但显然不行。（别担心，这只是完全错误，但这个思路展示了我想要的方向）

提前谢谢你！

Answer 1

你还有一些问题需要解决，特别是关于优雅性和其他更重要的事情。首先，你的用户会反感必须输入一个0，而你却没有对这个输入做任何有意义的处理。实际上，你还得专门写代码来忽略它！在字典里记录辅音的数量也显得不太优雅。你没有检查用户是否输入了所有字母，也没有处理大写字母。

下面是一段解决这些问题的代码，还有一些冗余的地方：

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

顺便提一下，你可能想把“ocurrances”改成“occurrences”。另外，如果元音的数量少于2，输出的结果就不太好看了。

Answer 2

另一种写代码的方法可能是这样的：

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"

你可以通过把搜索的部分提取出来，让这段代码更简洁一些：

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"