如果我有这样的字典：
{'avglen': 4.419354838709677, 'count:': 93, 'mosts:': 'your', 'longs:': ['stretched'], 'shorts:': ['i', 'a'],}
如何在新行上打印每个键值？
很长，但这里是我用来获取这本词典的代码。我几乎将每个键值添加到字典中。所以我想出了最短的单词，然后我将其添加到字典中。我也注意到有额外的冒号。但我想到了它的一部分值，我可以使用.replace（）来取走它？
def build_report(freq):
report={}
freq_list=list(freq.keys())
keys=sorted(freq, key=len)
#for shorts:
shortest=keys[0]
shortest = [keys[0]]
for key in keys[1:]:
if len(key) == len(shortest[0]):
shortest.append(key)
else:
break
report["shorts:"]=shortest
#for longs:
longest=keys[-1]
longest = [keys[-1]]
for key in reversed(keys[:-1]):
if len(key) == len(longest[0]):
longest.append(key)
else:
break
report["longs:"]=longest
#for mode:
val_list=list(freq.values())
sorted_val_list=sorted(val_list)
mode=sorted_val_list[-1]
for key in freq.keys():
if freq[key]==mode:
mode_word=key
report["mosts:"]=mode_word
# for word count:
wordcount=len(list(freq.keys()))
report["count:"]=wordcount
#for average length:
avg=list(freq.keys())
average=sum(map(len,avg))/len(avg)
report["avglen"]=average
#for freq dictionary of word to word count
freqs=freq
report["freqs:"]=freqs
return report
答案 0 :(得分：21)
如果你真的不想导入pprint但希望它“看起来像”字典，你可以这样做：
print("{" + "\n".join("{}: {}".format(k, v) for k, v in d.items()) + "}")
答案 1 :(得分：8)
您可能正在寻找漂亮的打印机标准库pprint。例如：
import pprint
pprint.pprint({'avglen': 4.419354838709677,
'count:': 93,
'mosts:': 'your',
'longs:': ['stretched'],
'shorts:': ['i', 'a'],})
输出
{'avglen': 4.419354838709677,
'count:': 93,
'longs:': ['stretched'],
'mosts:': 'your',
'shorts:': ['i', 'a']}
答案 2 :(得分：3)
迭代dict.items并打印：
>>> d = {'avglen': 4.419354838709677, 'count:': 93, 'mosts:': 'your', 'longs:': ['stretched'], 'shorts:': ['i', 'a'],}
>>> for k, v in d.items():
... print (k, '-->', v)
...
mosts: --> your
count: --> 93
avglen --> 4.41935483871
shorts: --> ['i', 'a']
longs: --> ['stretched']
或使用pprint模块：
>>> import pprint
>>> pprint.pprint(d)
{'avglen': 4.419354838709677,
'count:': 93,
'longs:': ['stretched'],
'mosts:': 'your',
'shorts:': ['i', 'a']}
答案 3 :(得分：1)
在过去的六个问题中，您似乎将这个可怜的字典用作各种文本索引对象。为什么不把它变成一个合适的班级？
from collections import Counter
textfreq = {
'I': 1, 'heaven': 1, 'filled': 1, 'their': 1, 'termed': 1, 'of': 4,
'And': 3, 'parts': 1, 'neer': 1, 'to': 2, 'song': 1, 'poets': 1,
'The': 1, 'a': 2, 'were': 2, 'verse': 1, 'your': 6, 'knows': 1,
'not': 1, 'half': 1, 'number': 1, 'but': 1, 'yours': 1, 'come': 2,
'rage': 1, 'age': 2, 'Though': 1, 'men': 1, 'fresh': 1, 'heavenly': 1,
'say': 1, 'alive': 1, 'truth': 1, 'this': 1, 'If': 2, 'than': 1,
'old': 1, 'believe': 1, 'Which': 1, 'that': 1, 'You': 1, 'faces': 1,
'yet': 1, 'poet': 1, 'in': 4, 'life': 1, 'most': 1, 'earthly': 1,
'will': 1, 'hides': 1, 'my': 3, 'papers': 1, 'is': 1, 'stretched': 1,
'rights': 1, 'eyes': 1, 'it': 3, 'yellowed': 1, 'Such': 1, 'So': 1,
'all': 1, 'lies': 1, 'the': 1, 'an': 1, 'as': 1, 'write': 1,
'child': 1, 'deserts': 1, 'shows': 1, 'tongue': 1, 'twice': 1,
'Be': 1, 'high': 1, 'some': 1, 'could': 1, 'should': 2, 'and': 2,
'touched': 1, 'like': 1, 'would': 1, 'Who': 1, 'tomb': 1, 'numbers': 1,
'antique': 1, 'scorned': 1, 'metre': 1, 'time': 2, 'touches': 1,
'be': 1, 'with': 2, 'true': 1, 'beauty': 1, 'rhyme': 1, 'less': 1,
'But': 1, 'graces': 1, 'live': 1
}
class TextStats():
def __init__(self, text=''):
if hasattr(text, 'wordfreq'):
# copy an existing TextStats object
self.wordfreq = Counter(text.wordfreq)
elif hasattr(text, 'keys'):
# load from an existing dict or Counter
self.wordfreq = Counter(text)
else:
# parse from a string
# TO DO - strip all punctuation
self.wordfreq = Counter(w for w in text.lower().split() if w)
@classmethod
def from_file(cls, fname):
with open(fname) as inf:
text = ' '.join(line.strip() for line in inf.readlines())
return cls(text.translate(None, '`~!@#$\'"'))
def __add__(self, otherTextStats):
return TextStats(self.wordfreq + otherTextStats.wordfreq)
def __str__(self):
return(
"Count: {}\n"
"Average len: {:0.4f}\n"
"Shortest: {}\n"
"Most common: {}\n"
"Longest: {}\n".format(
self.total_words,
self.average_word_length,
self.shortest_words,
self.most_common_words,
self.longest_words
)
)
@property
def unique_words(self):
return len(self.wordfreq)
@property
def total_words(self):
return sum(self.wordfreq.values())
@property
def total_letters(self):
return sum(len(w)*c for w,c in self.wordfreq.items())
@property
def average_word_length(self):
return float(self.total_letters) / self.total_words
@property
def shortest_words(self):
minlen = len(min(self.wordfreq, key=len))
return sorted(w for w in self.wordfreq if len(w)==minlen)
@property
def most_common_words(self):
most_common = self.wordfreq.most_common()
howmany = most_common[0][1] if most_common else 0
return sorted(w for w,c in most_common if c == howmany)
@property
def longest_words(self):
maxlen = len(max(self.wordfreq, key=len))
return sorted(w for w in self.wordfreq if len(w)==maxlen)
def main():
t = TextStats(textfreq)
u = TextStats.from_file('corpus.txt')
v = t + u
print(t)
print()
print(u)
print()
print(v)
if __name__=="__main__":
main()
答案 4 :(得分：0)
pprint的人来说，想知道为什么词典仍然不将每个条目打印到新行：可能是词典条目或整个词典太短了[em]。在这种情况下，请直接调用PrettyPrinter类，并按照docs中的指定，相应地设置width参数：
import pprint
stuff = {'avglen': 4.41, 'count:': 93, 'shorts:': ['i', 'a']}
pretty = pprint.PrettyPrinter(width=30)
pretty.pprint(stuff)
答案 5 :(得分：0)
要进行快速打印，可以使用字符串替换将换行符放入输出中。如果字典中包含列表，则此方法不会产生漂亮的结果；这些列表还将获得换行符。
td = {'avglen': 4.419354838709677, 'count:': 93, 'mosts:': 'your', 'longs:': ['stretched'], 'shorts:': ['i', 'a'],}
print(str(td).replace(', ',',\n '))
输出
{'avglen': 4.419354838709677,
'count:': 93,
'mosts:': 'your',
'longs:': ['stretched'],
'shorts:': ['i',
'a']}