Python 如何使字典的每个键值打印在新行上?
如果我有一本这样的字典:Python 如何使字典的每个键值打印在新行上?,python,python-3.x,dictionary,Python,Python 3.x,Dictionary,如果我有一本这样的字典: {'avglen': 4.419354838709677, 'count:': 93, 'mosts:': 'your', 'longs:': ['stretched'], 'shorts:': ['i', 'a'],} 如何使每个键值打印在新行上 很长,但这是我用来获取这本词典的代码。我几乎把每个键值都添加到了字典中。所以我算出了最短的单词,然后把它加到字典里。我也注意到有额外的冒号。但是我计算了它的一部分值,我可以使用.replace()将其删除 def buil
{'avglen': 4.419354838709677, 'count:': 93, 'mosts:': 'your', 'longs:': ['stretched'], 'shorts:': ['i', 'a'],}
如何使每个键值打印在新行上
很长,但这是我用来获取这本词典的代码。我几乎把每个键值都添加到了字典中。所以我算出了最短的单词,然后把它加到字典里。我也注意到有额外的冒号。但是我计算了它的一部分值,我可以使用.replace()将其删除
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
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']
或使用模块:
您可能正在寻找漂亮的打印机标准库。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']}
如果确实不想导入pprint,但希望它“看起来”像一本词典,可以执行以下操作:
print("{" + "\n".join("{!r}: {!r},".format(k, v) for k, v in d.items()) + "}")
在您过去的六个问题中,您似乎将这本糟糕的字典用作各种文本索引对象。为什么不让它成为一个合适的班级呢
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()
对于那些使用
pprint
的用户来说,他们想知道为什么词典仍然无法将每个条目打印到新行:可能是因为词典条目或整个词典太短。在这种情况下,直接调用PrettyPrinter
类,并相应地设置width
参数,如下所示:
为了快速打印,可以使用stringreplace在输出中添加换行符。如果字典包含列表,此方法不会给出漂亮的结果;这些列表也会有新行
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']}
请注意,键中似乎有剩余的冒号。我会看看你在做什么来生成字典。@DSM我添加了生成字典的代码。有没有一种方法可以不用导入任何东西就完成它?。第一个不是作为字典返回的,谢谢!
td = {'avglen': 4.419354838709677, 'count:': 93, 'mosts:': 'your', 'longs:': ['stretched'], 'shorts:': ['i', 'a'],}
print(str(td).replace(', ',',\n '))
{'avglen': 4.419354838709677,
'count:': 93,
'mosts:': 'your',
'longs:': ['stretched'],
'shorts:': ['i',
'a']}