Question

如何输出如下结果：

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。我想计算每个用户I，H和R的数量。请注意，在此特定情况下，我无法使用groupby中的itertools。在此先感谢您的帮助。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

此外：

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：

用户名称在L个字符

中都是合理的

在我的代码中，为了避免Sebastian代码中的复杂性，I，R，H的列在相同数量的LL字符中是合理的，这是此列中存在的所有结果的最大值

Answer 3

好吧，无论如何，使用groupby来解决这个问题毫无意义。对于初学者来说，您的数据没有排序（groupby没有为您排序组），而且行非常简单。

在处理每一行时只需保持计数。我假设你不知道你会得到什么标志：

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)

在此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)

上面的所有代码都是未经测试的，但应该大致有效。

Answer 4

这是一个使用嵌套dicts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]))

实施例

$ 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使用元组(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]))

使用和输出相同。

Answer 5

作为提示：

使用嵌套字典结构来计算出现次数：

用户 - ＆gt;字符 - ＆gt;出现用户角色

编写解析器代码并递增计数器并打印结果取决于你...一个很好的锻炼。

如何按项目计算每个用户组中的项目数

5 个答案:

实施例