如何计算每个用户组中的项目数

Question

我怎样才能输出这样的结果：

user    I   R   H
=================
atl001  2   1   0
cms017  1   2   1
lhc003  0   1   2

从这样的列表：

atl001 I
atl001 I
cms017 H
atl001 R
lhc003 H
cms017 R
cms017 I
lhc003 H
lhc003 R
cms017 R

即我想计算每个 I、H 和 R 的数量用户。请注意，在这种特殊情况下，我无法使用 itertools 中的 groupby 。预先感谢您的帮助。干杯！！

Answer 1

data='''atl001 I
atl001 I
cms017 H
atl001 R
lhc003 H
cms017 R
cms017 I
lhc003 H
lhc003 R
cms017 R'''

stats={}
for i in data.split('\n'):
    user, irh = i.split()
    u = stats.setdefault(user, {})
    u[irh] = u.setdefault(irh, 0) + 1

print 'user  I  R  H'
for user in sorted(stats):
    stat = stats[user]
    print user, stat.get('I', 0), stat.get('R', 0), stat.get('H', 0) 

Answer 2

data = 112*'cms017 R\n'

data = data + '''atl001 I
cms017 R
atl001 I
cms017 H
atl001 R
lhcabc003 H
cms017 R
lhcabc003 H
lhcabc003 R
cms017 R
cms017 R
cms017 R'''
print data,'\n'

stats = {}
d = {'I':0,'R':1,'H':2}
L = 0
for line in data.splitlines():
    user,irh = line.split()
    stats.setdefault(user,[0,0,0])
    stats[user][d[irh]] += 1
    L = max(L, len(user))

LL = len(str(max(max(stats[user])
                 for user in stats )))

cale = ' %%%ds %%%ds %%%ds' % (LL,LL,LL)
ch = 'user'.ljust(L) + cale % ('I','R','H')

print '%s\n%s' % (ch, len(ch)*'=')
print '\n'.join(user.ljust(L) + cale % tuple(stats[user])
                for user in sorted(stats.keys()))

结果

user        I   R   H
=====================
atl001      2   1   0
cms017      0 117   1
lhcabc003   0   1   2

。

另外：

data = 14*'cms017 R\n'

data = data + '''atl001 I
cms017 R
atl001 I
cms017 H
atl001 R
lhcabc003 H
cms017 R
lhcabc003 H
lhcabc003 R
cms017 R
cms017 R
cms017 R'''
print data,'\n'

Y = {}
L = 0
for line in data.splitlines():
    user,irh = line.split()
    L = max(L, len(user))
    if (user,irh) not in Y:
        Y.update({(user,'I'):0,(user,'R'):0,(user,'H'):0})
    Y[(user,irh)] += 1

LL = len(str(max(x for x in Y.itervalues())))

cale = '%%-%ds %%%ds %%%ds %%%ds' % (L,LL,LL,LL)
ch = cale % ('user','I','R','H')

print '%s\n%s' % (ch, len(ch)*'=')
li = sorted(Y.keys())
print '\n'.join(cale % (a[0],Y[b],Y[c],Y[a])
                for a,b,c in (li[x:x+3] for x in xrange(0,len(li),3)))

结果

user       I  R  H
==================
atl001     2  1  0
cms017     0 19  1
lhcabc003  0  1  2

。

PS：

用户的名字都是在L个字符中对齐的

在我的代码中，为了避免塞巴斯蒂安代码中的复杂性，I，R，H在相同数量的字符LL中对齐，这是最大的此列中出现的所有结果

Answer 3

好吧，无论如何，使用groupby来解决这个问题是没有意义的。首先，您的数据未排序（groupby 不会为您对组进行排序），并且这些行非常简单。

处理每一行时，只需记数即可。我假设您不知道会得到什么标志：

from sets import Set as set # python2.3 compatibility
counts = {} # counts stored in user -> dict(flag=counter) nested dicts
flags = set()
for line in inputfile:
    user, flag = line.strip().split()
    usercounts = counts.setdefault(user, {})
    usercounts[flag] = usercounts.setdefault(flag, 0) + 1
    flags.add(flag)

之后打印信息是迭代计数结构的问题。我假设用户名总是 6 个字符长：

flags = list(flags)
flags.sort()
users = counts.keys()
users.sort()
print "user  %s" % ('  '.join(flags))
print "=" * (6 + 3 * len(flags))
for user in users:
    line = [user]
    for flag in flags:
        line.append(counts[user].get(flag, 0))
    print '  '.join(line)

上面的所有代码都未经测试，但应该大致可以工作。

Answer 4

这是一个使用嵌套字典来计算作业状态并在打印前计算最大字段宽度的变体：

#!/usr/bin/env python
import fileinput
from sets import Set as set # python2.3

# parse job statuses
counter = {}
for line in fileinput.input():
    user, jobstatus = line.split()
    d = counter.setdefault(user, {})
    d[jobstatus] = d.setdefault(jobstatus, 0) + 1

# print job statuses
# . find field widths
status_names = set([name for st in counter.itervalues() for name in st])
maxstatuslens = [max([len(str(i)) for st in counter.itervalues()
                      for n, i in st.iteritems()
                      if name == n])
                 for name in status_names]
maxuserlen = max(map(len, counter))
row_format = (("%%-%ds " % maxuserlen) +
              " ".join(["%%%ds" % n for n in maxstatuslens]))
# . print header
header = row_format % (("user",) + tuple(status_names))
print header
print '='*len(header)
# . print rows
for user, statuses in counter.iteritems():
    print row_format % (
        (user,) + tuple([statuses.get(name, 0) for name in status_names]))

示例

$ python print-statuses.py <input.txt
user   I H R
============
lhc003 0 2 1
cms017 1 1 2
atl001 2 0 1

这是一个使用平面字典并以元组 (user, status_name) 作为键的变体：

#!/usr/bin/env python
import fileinput
from sets import Set as set # python 2.3

# parse job statuses
counter = {}
maxstatuslens = {}
maxuserlen = 0
for line in fileinput.input():
    key = user, status_name = tuple(line.split())
    i = counter[key] = counter.setdefault(key, 0) + 1
    maxstatuslens[status_name] = max(maxstatuslens.setdefault(status_name, 0),
                                     len(str(i)))
    maxuserlen = max(maxuserlen, len(user))

# print job statuses
row_format = (("%%-%ds " % maxuserlen) +
              " ".join(["%%%ds" % n for n in maxstatuslens.itervalues()]))
# . print header
header = row_format % (("user",) + tuple(maxstatuslens))
print header
print '='*len(header)
# . print rows
for user in set([k[0] for k in counter]):
    print row_format % ((user,) +
        tuple([counter.get((user, status), 0) for status in maxstatuslens]))

用法和输出是一样的。

Answer 5

提示：

使用嵌套字典结构来计算出现次数：

user ->字符-> 的字符的出现

用户编写解析器代码并递增计数器并打印结果
取决于你......一个很好的练习。