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Programming版 - Python问题请教
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进入Programming版参与讨论
1 (共1页)
x*z
发帖数: 1010
1
应该很简单,自己也能想出解决办法,但是肯定不是最优的,想找个
最优解。问题就是展开如下的range,变成一个sequence。
[0-3,4,8-9],13,[102-103,150],1000,...
-》
[0,1,2,3,4,8,9,13,102,103,150,1000,...]
我现在的想法就是先去匹配[],把[]全提出来后再按','去split,
然后再分别处理[]里面的内容,这样太复杂了,有什么简单办法么?
X****r
发帖数: 3557
2
What []s are for? i.e. what's the difference between
[1,2],3 and 1,2,3?

【在 x*z 的大作中提到】
: 应该很简单,自己也能想出解决办法,但是肯定不是最优的,想找个
: 最优解。问题就是展开如下的range,变成一个sequence。
: [0-3,4,8-9],13,[102-103,150],1000,...
: -》
: [0,1,2,3,4,8,9,13,102,103,150,1000,...]
: 我现在的想法就是先去匹配[],把[]全提出来后再按','去split,
: 然后再分别处理[]里面的内容,这样太复杂了,有什么简单办法么?

D*****r
发帖数: 6791
3
应该是[]里的内容可以带-,外面没有这种。
其实递归做很好做,但是如果输入当字符串处理,递归时候,传字符串岂不是复制了大
量的中间结果了?
要么就一个数一个数的读,遇到‘-’就进栈,然后出栈的时候,一口气出到截止的数
。直到出完。

【在 X****r 的大作中提到】
: What []s are for? i.e. what's the difference between
: [1,2],3 and 1,2,3?

M*P
发帖数: 6456
4
there is no difference in the example,just remove [] and split by , and
match -

【在 X****r 的大作中提到】
: What []s are for? i.e. what's the difference between
: [1,2],3 and 1,2,3?

X****r
发帖数: 3557
5
It is very easy if we just ignore '[]'s:
import re
re.sub('(\d+)-(\d+)', lambda m: ','.join(map(str, range(int(m.group(1)), int
(m.group(2))+1))), '[0-3,4,8-9],13,[102-103,150],1000', 0).replace('[', '').
replace(']', '')

【在 M*P 的大作中提到】
: there is no difference in the example,just remove [] and split by , and
: match -

l********a
发帖数: 1154
6

int
).
顶这个,very pythonic
re,lambda,map齐活儿了

【在 X****r 的大作中提到】
: It is very easy if we just ignore '[]'s:
: import re
: re.sub('(\d+)-(\d+)', lambda m: ','.join(map(str, range(int(m.group(1)), int
: (m.group(2))+1))), '[0-3,4,8-9],13,[102-103,150],1000', 0).replace('[', '').
: replace(']', '')

P********e
发帖数: 2610
7
我很好奇,你怎么什么语言都熟阿?

int
).

【在 X****r 的大作中提到】
: It is very easy if we just ignore '[]'s:
: import re
: re.sub('(\d+)-(\d+)', lambda m: ','.join(map(str, range(int(m.group(1)), int
: (m.group(2))+1))), '[0-3,4,8-9],13,[102-103,150],1000', 0).replace('[', '').
: replace(']', '')

X****r
发帖数: 3557
8
哪里有。Python我只是过得去而已。一个程序员总要写些脚本什么的。

【在 P********e 的大作中提到】
: 我很好奇,你怎么什么语言都熟阿?
:
: int
: ).

L*******r
发帖数: 1011
9
:D 你的过得去还是比大部分人强的.

【在 X****r 的大作中提到】
: 哪里有。Python我只是过得去而已。一个程序员总要写些脚本什么的。
v*s
发帖数: 946
10
看懂了,受教了,原来re.sub 可以带lambda进去玩, 牛!!

int
).

【在 X****r 的大作中提到】
: It is very easy if we just ignore '[]'s:
: import re
: re.sub('(\d+)-(\d+)', lambda m: ','.join(map(str, range(int(m.group(1)), int
: (m.group(2))+1))), '[0-3,4,8-9],13,[102-103,150],1000', 0).replace('[', '').
: replace(']', '')

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进入Programming版参与讨论
j*****k
发帖数: 1198
11
How to convert [1,2,3,4,5,6, 7,10, 15,17,19,21] to
'1-8, 10, 15-23-2'?
Thanks
p**o
发帖数: 3409
12

def parse( s ) :

result = ''
lst = [ eval(i) for i in s[1:-1].split(', ') ]
head, p0 = lst[0], lst[1]
diff0 = p0 - head
flag = True
num_consecutives = 2

for p in lst[2:] :
if not flag :
flag = True
diff0 = p - p0
p0 = p
continue
diff = p - p0
if flag and diff0 <> diff :
if 2 == num_consecutives :
ss = str(head)
head = p0
diff0 = p - p0
else:
ss = str(head) + '-' + str(p0)
if 1 <> diff0 :
ss += '-' + str(diff0)
head = p
flag = False
num_consecutives = 2
result += ss + ', '
else:
num_consecutives += 1
p0 = p
#end_for

if 2 == num_consecutives :
ss = str(head)
else:
ss = str(head) + '-' + str(p0)
if 1 <> diff0 :
ss += '-' + str(diff0)
result += ss
return result
if __name__ == '__main__' :
ss = ['[1, 2, 3, 4, 5, 6, 7, 10, 15, 17, 19, 21]',\
'[1, 3, 4, 5, 6, 7, 10, 15, 17, 19, 21, 24]',\
'[1, 3, 5, 6, 7, 10, 12, 15, 17, 19, 21, 24, 27, 30]' ]
for s in ss :
print( s )
print( parse( s ) )
print
[1, 2, 3, 4, 5, 6, 7, 10, 15, 17, 19, 21]
1-7, 10, 15-21-2
[1, 3, 4, 5, 6, 7, 10, 15, 17, 19, 21, 24]
1, 3-7, 10, 15-21-2, 24
[1, 3, 5, 6, 7, 10, 12, 15, 17, 19, 21, 24, 27, 30]
1-5-2, 6, 7, 10, 12, 15-21-2, 24-30-3

【在 j*****k 的大作中提到】
: How to convert [1,2,3,4,5,6, 7,10, 15,17,19,21] to
: '1-8, 10, 15-23-2'?
: Thanks

x*z
发帖数: 1010
13
应该很简单,自己也能想出解决办法,但是肯定不是最优的,想找个
最优解。问题就是展开如下的range,变成一个sequence。
[0-3,4,8-9],13,[102-103,150],1000,...
-》
[0,1,2,3,4,8,9,13,102,103,150,1000,...]
我现在的想法就是先去匹配[],把[]全提出来后再按','去split,
然后再分别处理[]里面的内容,这样太复杂了,有什么简单办法么?
X****r
发帖数: 3557
14
What []s are for? i.e. what's the difference between
[1,2],3 and 1,2,3?

【在 x*z 的大作中提到】
: 应该很简单,自己也能想出解决办法,但是肯定不是最优的,想找个
: 最优解。问题就是展开如下的range,变成一个sequence。
: [0-3,4,8-9],13,[102-103,150],1000,...
: -》
: [0,1,2,3,4,8,9,13,102,103,150,1000,...]
: 我现在的想法就是先去匹配[],把[]全提出来后再按','去split,
: 然后再分别处理[]里面的内容,这样太复杂了,有什么简单办法么?

D*****r
发帖数: 6791
15
应该是[]里的内容可以带-,外面没有这种。
其实递归做很好做,但是如果输入当字符串处理,递归时候,传字符串岂不是复制了大
量的中间结果了?
要么就一个数一个数的读,遇到‘-’就进栈,然后出栈的时候,一口气出到截止的数
。直到出完。

【在 X****r 的大作中提到】
: What []s are for? i.e. what's the difference between
: [1,2],3 and 1,2,3?

M*P
发帖数: 6456
16
there is no difference in the example,just remove [] and split by , and
match -

【在 X****r 的大作中提到】
: What []s are for? i.e. what's the difference between
: [1,2],3 and 1,2,3?

X****r
发帖数: 3557
17
It is very easy if we just ignore '[]'s:
import re
re.sub('(\d+)-(\d+)', lambda m: ','.join(map(str, range(int(m.group(1)), int
(m.group(2))+1))), '[0-3,4,8-9],13,[102-103,150],1000', 0).replace('[', '').
replace(']', '')

【在 M*P 的大作中提到】
: there is no difference in the example,just remove [] and split by , and
: match -

l********a
发帖数: 1154
18

int
).
顶这个,very pythonic
re,lambda,map齐活儿了

【在 X****r 的大作中提到】
: It is very easy if we just ignore '[]'s:
: import re
: re.sub('(\d+)-(\d+)', lambda m: ','.join(map(str, range(int(m.group(1)), int
: (m.group(2))+1))), '[0-3,4,8-9],13,[102-103,150],1000', 0).replace('[', '').
: replace(']', '')

P********e
发帖数: 2610
19
我很好奇,你怎么什么语言都熟阿?

int
).

【在 X****r 的大作中提到】
: It is very easy if we just ignore '[]'s:
: import re
: re.sub('(\d+)-(\d+)', lambda m: ','.join(map(str, range(int(m.group(1)), int
: (m.group(2))+1))), '[0-3,4,8-9],13,[102-103,150],1000', 0).replace('[', '').
: replace(']', '')

X****r
发帖数: 3557
20
哪里有。Python我只是过得去而已。一个程序员总要写些脚本什么的。

【在 P********e 的大作中提到】
: 我很好奇,你怎么什么语言都熟阿?
:
: int
: ).

相关主题
请教一个parser的问题请教改numpy array的dtype
a simple question[合集] First impression on Python
A question about sharing data inside a C++ class这问题有没有好办法做?
进入Programming版参与讨论
L*******r
发帖数: 1011
21
:D 你的过得去还是比大部分人强的.

【在 X****r 的大作中提到】
: 哪里有。Python我只是过得去而已。一个程序员总要写些脚本什么的。
v*s
发帖数: 946
22
看懂了,受教了,原来re.sub 可以带lambda进去玩, 牛!!

int
).

【在 X****r 的大作中提到】
: It is very easy if we just ignore '[]'s:
: import re
: re.sub('(\d+)-(\d+)', lambda m: ','.join(map(str, range(int(m.group(1)), int
: (m.group(2))+1))), '[0-3,4,8-9],13,[102-103,150],1000', 0).replace('[', '').
: replace(']', '')

j*****k
发帖数: 1198
23
How to convert [1,2,3,4,5,6, 7,10, 15,17,19,21] to
'1-8, 10, 15-23-2'?
Thanks
p**o
发帖数: 3409
24

def parse( s ) :

result = ''
lst = [ eval(i) for i in s[1:-1].split(', ') ]
head, p0 = lst[0], lst[1]
diff0 = p0 - head
flag = True
num_consecutives = 2

for p in lst[2:] :
if not flag :
flag = True
diff0 = p - p0
p0 = p
continue
diff = p - p0
if flag and diff0 <> diff :
if 2 == num_consecutives :
ss = str(head)
head = p0
diff0 = p - p0
else:
ss = str(head) + '-' + str(p0)
if 1 <> diff0 :
ss += '-' + str(diff0)
head = p
flag = False
num_consecutives = 2
result += ss + ', '
else:
num_consecutives += 1
p0 = p
#end_for

if 2 == num_consecutives :
ss = str(head)
else:
ss = str(head) + '-' + str(p0)
if 1 <> diff0 :
ss += '-' + str(diff0)
result += ss
return result
if __name__ == '__main__' :
ss = ['[1, 2, 3, 4, 5, 6, 7, 10, 15, 17, 19, 21]',\
'[1, 3, 4, 5, 6, 7, 10, 15, 17, 19, 21, 24]',\
'[1, 3, 5, 6, 7, 10, 12, 15, 17, 19, 21, 24, 27, 30]' ]
for s in ss :
print( s )
print( parse( s ) )
print
[1, 2, 3, 4, 5, 6, 7, 10, 15, 17, 19, 21]
1-7, 10, 15-21-2
[1, 3, 4, 5, 6, 7, 10, 15, 17, 19, 21, 24]
1, 3-7, 10, 15-21-2, 24
[1, 3, 5, 6, 7, 10, 12, 15, 17, 19, 21, 24, 27, 30]
1-5-2, 6, 7, 10, 12, 15-21-2, 24-30-3

【在 j*****k 的大作中提到】
: How to convert [1,2,3,4,5,6, 7,10, 15,17,19,21] to
: '1-8, 10, 15-23-2'?
: Thanks

w****i
发帖数: 964
25
def parse(s):
d2 = [1, 1] + [s[i]*2-s[i-1]-s[i+1] for i in xrange(1, len(s)-1)] + [1,
1]
outstr = ''
for i in xrange(len(s)):
if d2[i+1] == 0: continue
if d2[i] != 0 and d2[i+2] != 0: outstr += ', %d' % s[i]
if d2[i] == 0:
outstr += ', %d-%d' % (s[start],s[i])
if s[i] - s[i-1] > 1: outstr += '-%d' % (s[i]-s[i-1])
d2[i+2] = 1
if d2[i+2] == 0: start = i
return outstr[2:]
w****i
发帖数: 964
26
def parse(s):
d2 = [1, 1] + [s[i]*2-s[i-1]-s[i+1] for i in xrange(1, len(s)-1)] + [1,
1]
outstr = ''
for i in xrange(len(s)):
if d2[i+1] == 0: continue
if d2[i] != 0 and d2[i+2] != 0: outstr += ', %d' % s[i]
if d2[i] == 0:
outstr += ', %d-%d' % (s[start],s[i])
if s[i] - s[i-1] > 1: outstr += '-%d' % (s[i]-s[i-1])
d2[i+2] = 1
if d2[i+2] == 0: start = i
return outstr[2:]
w****i
发帖数: 964
27
def parse(s):
d2 = [1, 1] + [s[i]*2-s[i-1]-s[i+1] for i in xrange(1, len(s)-1)] + [1,
1]
outstr = ''
for i in xrange(len(s)):
if d2[i+1] == 0: continue
if d2[i] != 0 and d2[i+2] != 0: outstr += ', %d' % s[i]
if d2[i] == 0:
outstr += ', %d-%d' % (s[start],s[i])
if s[i] - s[i-1] > 1: outstr += '-%d' % (s[i]-s[i-1])
d2[i+2] = 1
if d2[i+2] == 0: start = i
return outstr[2:]
j*****k
发帖数: 1198
28
Both solutions are very good. In westxi's solution, it would be better if
change %d to %s, s[i] - s[i-1] > 1 to s[i] - s[i-1] != 1
Thanks, very much
j*****k
发帖数: 1198
29
Pulo, if the array is [1, 3, 5, 6, 7, 10, 12, 15, 17, 19, 21, 24, 27, 30, 8,
10], your code will have a problem to output the last one
j*****k
发帖数: 1198
30
Here is another version:
def parse(m_Array):
m_Out = '%s'%m_Array[0]
m_Flag = 0
m_Array = m_Array + [1]
for i in range(1,len(m_Array)-1):
if (m_Array[i] - m_Array[i - 1]) == (m_Array[i + 1] - m_Array[i]):
m_Flag += 1
if m_Flag == 0:
m_Out += '%s'%m_Array[i]
else:
if m_Flag >= 1:
m_Out += ':%s'% m_Array[i]
m_Step = m_Array[i] - m_Array[i-1]
if m_Step <> 1:
m_Out += ':%s'%m_Step
if i<>len(m_Array)-2: m_Out += ', '
m_Flag = -1
elif m_Flag ==0:
m_Out += ", %s"% m_Array[i]
elif m_Flag == -1:
m_Out += "%s"% m_Array[i]
m_Flag = 0
return m_Out
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p**o
发帖数: 3409
31
我假设输入的list是排过序的,而且没有重复元素
你可以先把list转成set再转回list,完成排重和排序
lst = [1, 3, 5, 6, 7, 10, 12, 15, 17, 19, 21, 24, 27, 30, 8, 10]
lst = list( set( lst ) )
输出
[1, 3, 5, 6, 7, 8, 10, 12, 15, 17, 19, 21, 24, 27, 30]
1-5-2, 6-8, 10, 12, 15-21-2, 24-30-3

8,

【在 j*****k 的大作中提到】
: Pulo, if the array is [1, 3, 5, 6, 7, 10, 12, 15, 17, 19, 21, 24, 27, 30, 8,
: 10], your code will have a problem to output the last one

1 (共1页)
进入Programming版参与讨论
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