python - 如何从动态键获取python中的字典值

Question

from xlrd import open_workbook

book = open_workbook('trial.xls')
sheet=book.sheet_by_index(0)

name_email={}

i=0
for row_index in range(sheet.nrows):
    if name_email.has_key(sheet.cell(row_index,i).value):
         name_email[str(sheet.cell(row_index,i).value.strip())]=sheet.cell(row_index,i+1).value,)
    else:              
        abc = str(sheet.cell(row_index,i).value.strip())
        print repr(abc)
        print '"{0}"'.format(repr(abc))
        # print name_email[abc]
        name_email[str(sheet.cell(row_index,i).value.strip())]=sheet.cell(row_index,i+1).value
i+=1            
print name_email.keys()
print name_email

输出是: 'manoj' “'manoj'” 'dibyendu' “'dibyendu'” '苏拉夫' “'sourav'” ['dibyendu', 'sourav', 'manoj'] {'dibyendu': (u'd.b@gmail.com',), 'sourav': (u's.b@gmail.com',), 'manoj': (u'm.c@gmail.com' ,)}

但仍然无法访问 print name_email[abc] 并抛出错误 打印 name_email[abc] KeyError: 'manoj'

Answer 1

它就像你想象的那样工作:

>>> mydict = {"spam": "flat", "eggs": "round"}
>>> print mydict["eggs"]
round
>>> mykey = "eggs"
>>> print mydict[mykey]
round
>>>

您的变量 abc 显然不包含您认为的内容，如果它在您对 key 进行硬编码时有效的话。