我显然是python的新手。所以我为我的赋值写的代码给了我正确的输出,然而,它只会伤害你的眼睛。所以我想知道是否有一种方法可以使代码更紧凑?
下面是我的代码:
import sys
period_count = 0
excl_count = 0
quest_count = 0
colon_count = 0
dquote_count = 0
comma_count = 0
scolon_count = 0
slash_count = 0
bslash_count = 0
lparen_count = 0
rparen_count = 0
quote_count = 0
for line in sys.stdin:
for char in line:
if char == ".":
period_count = period_count + 1
if char == "!":
excl_count = excl_count +1
if char == "?":
quest_count = quest_count +1
if char == ":":
colon_count = colon_count +1
if char == "\"":
dquote_count = dquote_count +1
if char == ",":
comma_count = comma_count +1
if char == ";":
scolon_count = scolon_count +1
if char == "/":
slash_count = slash_count +1
if char == "\\":
bslash_count = bslash_count +1
if char == "(":
lparen_count = lparen_count +1
if char == "(":
rparen_count = rparen_count +1
if char == "'":
quote_count = quote_count +1
print("{0} {1:>3} {0:>4} {2:>5} {0:>4}".format("|", "mark", "number"))
print("{0} {1} {0} {2} {0}".format("|", "_______", "_________"))
print("{0} {1:>3} {0:>5} {2:>5} {0:>5}".format("|", ".", period_count))
print("{0} {1:>3} {0:>5} {2:>5} {0:>5}".format("|", "!", excl_count))
print("{0} {1:>3} {0:>5} {2:>5} {0:>5}".format("|", "?", quest_count))
print("{0} {1:>3} {0:>5} {2:>5} {0:>5}".format("|", ":", colon_count))
print("{0} {1:>3} {0:>5} {2:>5} {0:>5}".format("|", "\"", dquote_count))
print("{0} {1:>3} {0:>5} {2:>5} {0:>5}".format("|", ",", comma_count))
print("{0} {1:>3} {0:>5} {2:>5} {0:>5}".format("|", ";", scolon_count))
print("{0} {1:>3} {0:>5} {2:>5} {0:>5}".format("|", "/", slash_count))
print("{0} {1:>3} {0:>5} {2:>5} {0:>5}".format("|", "\\", bslash_count))
print("{0} {1:>3} {0:>5} {2:>5} {0:>5}".format("|", "(", lparen_count))
print("{0} {1:>3} {0:>5} {2:>5} {0:>5}".format("|", ")", rparen_count))
print("{0} {1:>3} {0:>5} {2:>5} {0:>5}".format("|", "'", quote_count))任务是分别计算每个标点符号的数量,并将该标点符号和结果打印在表格中。
任何帮助都将不胜感激!
发布于 2017-10-06 20:53:05
您可以对字符串进行散列,并将其归功于tobias_k的理解:
dictionary = {c: 0 for c in ".!?:\",;/\\()'"}
for line in sys.stdin:
for char in line:
if char in dictionary:
dictionary[char] += 1
for char, counter in dictionary.items():
print(...) # formatted string goes here编辑
每次在评论中与Chris_Rands交谈时,如果任务是简单地计算所有标点符号,那么我建议稍微修改一下上面的内容:
import string
dictionary = {c: 0 for c in string.punctuation}Rest如下所示。在我的机器上,string.punctuation是'!"#$%&'()*+,-./:;<=>?@[\]^_{|}~' (根据文档,该集合是基于区域设置的)。
最终更新
感谢Muposat提醒我使用defaultdict:当0是int构造函数的默认值时,不需要初始化键和计数器:
import sys
import string
dictionary = defaultdict(int)
for line in sys.stdin:
for char in line:
if char in string.punctuation:
dictionary[char] += 1
for char, counter in dictionary.items():
print(...) # formatted string goes here发布于 2017-10-06 21:13:38
这个较短的版本将使用内置类collections.Counter计算所有字符频率,并打印所需的频率(包括原始表格格式):
import sys, collections
freqs = collections.Counter(char for line in sys.stdin.readlines() for char in line)
print("| mark | count |")
print("|_______|_______|")
for char in ".!?:\",;/\\()'":
print("| {:>5} | {:>5} |".format(char, freqs.get(char, 0)))或者同样的想法,只是为了好玩:
(lambda freqs=__import__('collections').Counter(char for line in __import__('sys').stdin.readlines() for char in line):print("| mark | count |\n|_______|_______|\n"+"".join("| {:>5} | {:>5} |\n".format(char, freqs.get(char, 0)) for char in ".!?:\",;/\\()'")))()https://stackoverflow.com/questions/46606133
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