使用 openpyxl 创建一个字典,其中键是 excel sheet 中的 header_name
Create a dictionary where key would be the header_name from excel sheet using openpyxl
from openpyxl import *
Variable_Model_Path=defaultdict(list)
Sheet_Name=wb.get_sheet_by_name(Sheet)
for row in Sheet_Name.iter_rows():
Row=[cell.value for cell in row]
for cell_ind,cell in enumerate(Row):
print cell
以上是我的示例代码。
我有一个 excel sheet,其中有 headers var1, var2,var3。我想创建一个字典 {Var1:1,4,7,10},其中 Var1 是 header 名称,值是对应的列值。我正在使用 openpyxl module.Here 变量名称和值将由用户提供,它们将是动态的。
Var1 Var2 Var3
- 1 2 3
- 4 5 6
- 7 8 9
- 10 11 12
这个:
from openpyxl import *
from collections import defaultdict
path = "/Users/romain/Desktop/Classeur2.xlsx"
Sheet = "Feuil1"
wb = load_workbook(path)
Variable_Model_Path = defaultdict(list)
Sheet_Name = wb.get_sheet_by_name(Sheet)
keys = False
for row in Sheet_Name.iter_rows():
if not keys:
keys = [cell.value for cell in row]
continue
Row=[cell.value for cell in row]
for cell_ind,cell in enumerate(Row):
Variable_Model_Path[keys[cell_ind]].append( cell)
给我:
Variable_Model_Path
defaultdict(list,
{u'var 1': [1L, 4L, 7L],
u'var 2': [2L, 5L, 8L],
u'var 3': [3L, 6L, 9L]})
from openpyxl import *
Variable_Model_Path=defaultdict(list)
Sheet_Name=wb.get_sheet_by_name(Sheet)
for row in Sheet_Name.iter_rows():
Row=[cell.value for cell in row]
for cell_ind,cell in enumerate(Row):
print cell
以上是我的示例代码。
我有一个 excel sheet,其中有 headers var1, var2,var3。我想创建一个字典 {Var1:1,4,7,10},其中 Var1 是 header 名称,值是对应的列值。我正在使用 openpyxl module.Here 变量名称和值将由用户提供,它们将是动态的。
Var1 Var2 Var3
- 1 2 3
- 4 5 6
- 7 8 9
- 10 11 12
这个:
from openpyxl import *
from collections import defaultdict
path = "/Users/romain/Desktop/Classeur2.xlsx"
Sheet = "Feuil1"
wb = load_workbook(path)
Variable_Model_Path = defaultdict(list)
Sheet_Name = wb.get_sheet_by_name(Sheet)
keys = False
for row in Sheet_Name.iter_rows():
if not keys:
keys = [cell.value for cell in row]
continue
Row=[cell.value for cell in row]
for cell_ind,cell in enumerate(Row):
Variable_Model_Path[keys[cell_ind]].append( cell)
给我:
Variable_Model_Path
defaultdict(list,
{u'var 1': [1L, 4L, 7L],
u'var 2': [2L, 5L, 8L],
u'var 3': [3L, 6L, 9L]})