循环遍历字典并以特定格式打印

Question

I have the following dictionary:

# a small DB of people who stole my books

dic = {
'Cohen'     : 'Calvino' 'Evertt' 'Borges',
'Larry'     : 'The Bible', 
'Volanski'  : 'Phone Book'
}

# here's an abortive attempt to print it in a CSV format
for k in dic.keys():
    print (k, '\t')
for v in dic.keys():
    print (dic[v], ' ')

This is the the ugly output:

Volanski    
Phone Book  
CalvinoEverttBorges  
The Bible  
Cohen   
Phone Book  
CalvinoEverttBorges  
The Bible  
Larry   
Phone Book  
CalvinoEverttBorges  
The Bible 

This is how I wished the output to look like:

Cohen      Larry       Volanski  
Calvino    The Bible   Phone Book  
Evertt  
Borgest  

(only tab-separated which I didn't manage to show here)

Answer 1

dic = {
   'Cohen'     : ['Calvino', 'Evertt', 'Borges'],
   'Larry'     : ['The Bible'],
   'Volanski'  : ['Phone Book']
}


ordered = []
maxlen = 0

for k in sorted(dic.keys()):
    lst = [k] + dic[k]
    maxlen = max(len(lst), maxlen)
    ordered.append(iter(lst))

for i in range(maxlen):
    print "\t".join(next(j, '') for j in ordered)

Answer 2

You can work out neater formatting

dic = {'Cohen'     : ['Calvino', 'Evertt', 'Borges'],
       'Larry'     : ['The Bible'],
       'Volanski'  : ['Phone Book']}

# Get max name size
mx_nm_len = len(max(dic,key=len))
mx_bk_len = max([len(max(books, key=len)) for books in dic.itervalues()])

# Store max name size + 1
mx = max([mx_nm_len, mx_bk_len]) + 1

# Store people
keys = dic.keys()

# Create generic format code to print neat list
fmat = ("%-"+str(mx)+"s")*len(keys)

# Print header line
print fmat % tuple(keys)

# similar to zip command but works for any number of lists
# Assumes all dic.values() are lists
# "zips" to longest list and uses None when any given list runs out of values
books = map(None, *dic.values())

# replaces None values in row outputs with empty strings and prints result using
# string format code (fmat)
for row in books:
    row = tuple([book if book!= None else "" for book in row])
    print fmat % row

Answer 3

The point is that you didn't define the data in the right way to begin with. The way you've defined the first entry is understood by Python in precisely the way it's printed: a single string. If you want multiple elements, you need to define your elements as containing multiple elements:

dic = {
  'Cohen'     : ['Calvino', 'Evertt', 'Borges'],
  'Larry'     : ['The Bible'], 
  'Volanski'  : ['Phone Book']
}

Now you can simply do this:

for key, value in dic.items():
    print "%s\t%s" % (key, "\t".join(value)) 

Edit OK, didn't see you wanted the names across the top and titles downwards. A bit trickier, but this will do it:

import itertools
print "\t".join(dic.keys())
for books in itertools.izip_longest(*dic.values(), fillvalue=''):
    print "\t".join(books)

Answer 4

This solution ensures the output is sorted and items wind up in the correct columns. I use expandtabs() to set the tab length, to make it pretty print.

import itertools

dic = {'Cohen':['Calvino', 'Evertt','Borges'],'Larry': ['The Bible'], 'Volanski'  :['Phone Book']}

sorted_keys = sorted(dic.keys())
for name in sorted_keys:
    print (name+'\t').expandtabs(15),
print
zippp = list(itertools.izip_longest(*[dic[i] for i in sorted_keys]))
for i in xrange(len(zippp[0])):
    for j in xrange(len(zippp)):
        if zippp[i][j]:
            print (zippp[i][j]+'\t').expandtabs(15),
    print

Produces:

Cohen           Larry           Volanski       
Calvino         The Bible       Phone Book     
Evertt         
Borges