Nach langer Suche nach Daten haben wir diesen Konflikt gelöst, den viele unserer Leser haben. Wir bieten Ihnen die Lösung und möchten Sie tatkräftig unterstützen.
Beispiel: xlrd select sheet by name
from __future__ import print_function
from os.path import join, dirname, abspath
import xlrd
fname = join(dirname(dirname(abspath(__file__))),'test_data','Cad Data Mar 2014.xlsx')# Open the workbook
xl_workbook = xlrd.open_workbook(fname)# List sheet names, and pull a sheet by name#
sheet_names = xl_workbook.sheet_names()print('Sheet Names', sheet_names)
xl_sheet = xl_workbook.sheet_by_name(sheet_names[0])# Or grab the first sheet by index # (sheets are zero-indexed)#
xl_sheet = xl_workbook.sheet_by_index(0)print('Sheet name: %s'% xl_sheet.name)# Pull the first row by index# (rows/columns are also zero-indexed)#
row = xl_sheet.row(0)# 1st row# Print 1st row values and types#from xlrd.sheet import ctype_text
print('(Column #) type:value')for idx, cell_obj inenumerate(row):
cell_type_str = ctype_text.get(cell_obj.ctype,'unknown type')print('(%s) %s %s'%(idx, cell_type_str, cell_obj.value))# Print all values, iterating through rows and columns#
num_cols = xl_sheet.ncols # Number of columnsfor row_idx inrange(0, xl_sheet.nrows):# Iterate through rowsprint('-'*40)print('Row: %s'% row_idx)# Print row numberfor col_idx inrange(0, num_cols):# Iterate through columns
cell_obj = xl_sheet.cell(row_idx, col_idx)# Get cell object by row, colprint('Column: [%s] cell_obj: [%s]'%(col_idx, cell_obj))Denken Sie daran, dass Sie die Fähigkeit zur Klärung haben.