使用 openpyxl 在 Excel sheet 中的文本字符串中搜索单词

Question

我正在尝试在具有如下文本字符串的单元格中搜索单词（能源；绿色建筑；高性能建筑）。这是我写的代码，我得到一个语法错误

for row in ws.iter_rows('D2:D11'):
    for cell in row:
        if 'Energy' in ws.cell.value :
            Print 'yes'

显然，我不想打印 yes，这是为了测试搜索功能。

此外，我想获取单元格位置，然后告诉openpyxl为E列下同一行中的单元格分配颜色。这是我的快照 Excel sheet. 我知道如何使用此命令分配颜色

c.fill = PatternFill(start_color='FFFFE0', end_color='FFFFE0' fill_type='solid')

我只需要帮助获取单元格位置（具有匹配文本的单元格）并将其行号分配给 E 列中的另一个单元格

更新：我在下面写了这段代码，对我来说工作正常：

import xml.etree.ElementTree as ET



fhand = open ('My_Collection')    
tree =ET.parse('My_Collection.xml')
data= fhand.read()
root = tree.getroot()
tree = ET.fromstring(data)

title_list= ['Title']
year_list = ['Year']
author_list= ['Author']
label_list = ['Label']



for child in tree:
    for children in child:
        if children.find('.//title')is None :
            t='N'
        else:
            t=children.find('.//title').text
        title_list.append(t)
    print title_list
    print len(title_list)


for child in tree:
    for children in child:
        if children.find('.//year')is None :
            y='N'
        else:
            y=children.find('.//year').text
        year_list.append(y)
    print year_list
    print len(year_list)


for child in tree:
    for children in child:
        if children.find('.//author')is None :
            a='N'
        else:
            a=children.find('.//author').text
        author_list.append(a)
    print author_list
    print len(author_list)


for child in tree:
    for children in child:
        if children.find('label')is None :
            l='N'
        else:
            l=children.find('label').text
        label_list.append(l)
    print label_list
print len(author_list) 





Modified_label_list=list()        
import re
for labels in label_list:

    all_labels=labels.split(';')

    for a_l in all_labels:
        if a_l not in  Modified_label_list: 
            Modified_label_list.append(a_l)
        else:
            continue
print Modified_label_list
print len(Modified_label_list)
label_list_for_col_header= Modified_label_list[1:]
print label_list_for_col_header
print len(label_list_for_col_header)




from openpyxl import Workbook 
wb = Workbook() 
ws = wb.active 


for row in zip(title_list, year_list, author_list, label_list): 
        ws.append(row)




r = 5
for N in label_list_for_col_header:
    ws.cell(row=1, column=r).value = str(N)
    r += 1


from openpyxl.styles import PatternFill 


general_lst= list()



COLOR_INDEX = ['FF000000', 'FFFFFFFF', 'FFFF0000', 'FF00FF00', 'FF0000FF',
               'FFFFFF00', 'FFFF00FF', 'FF00FFFF', 'FF800000', 'FF008000', 'FF000080',
               'FF808000', 'FF800080', 'FF008080', 'FFC0C0C0', 'FF808080', 'FF9999FF',
               'FF993366', 'FFFFFFCC', 'FFCCFFFF', 'FF660066', 'FFFF8080', 'FF0066CC',
               'FFCCCCFF', 'FF000080', 'FFFF00FF', 'FFFFFF00', 'FF00FFFF', 'FF800080',
               'FF800000', 'FF008080', 'FF0000FF', 'FF00CCFF', 'FFCCFFFF', 'FFCCFFCC',
               'FFFFFF99', 'FF99CCFF', 'FFFF99CC', 'FFCC99FF', 'FFFFCC99', 'FF3366FF',
               'FF33CCCC', 'FF99CC00', 'FFFFCC00', 'FFFF9900', 'FFFF6600', 'FF666699',
               'FF969696', 'FF003366', 'FF339966', 'FF003300', 'FF333300', 'FF993300',
               'FF993366', 'FF333399', 'FF333333']

import random
color_lst= random.sample(COLOR_INDEX, len(label_list_for_col_header))
print color_lst

print int(label_list_for_col_header.index(label_list_for_col_header[0]))

h= len(title_list)
m= 0    
for lbls in label_list_for_col_header: 
    j= int(label_list_for_col_header.index(lbls))+5
    for row in ws.iter_rows('D2:D11'):
        for cell in  row:

            if lbls in cell.value : 
                general_lst.append(cell.row)
                for items in range(len(general_lst)):

                    ws.cell(row = general_lst[items], column = j).fill = PatternFill(start_color=str(color_lst[m]), end_color=str(color_lst[m]) , fill_type='solid')
    general_lst = []
    m +=1       


ws.column_dimensions['A'].width = 70    
ws.column_dimensions['C'].width = 23
ws.column_dimensions['B'].width = 5        
wb.save("Test61.xlsx")

Answer 1

对于搜索方法，我建议您查看 answer to this question. Create a dictionary of terms to search in a single pass and use it as often as you like. However, you might also like to know that openpyxl also supports conditional formatting so that you can delegate the formatting to Excel. See the very bottom of the examples in the documentation。上周我自己第一次使用这个。 Excel API 和往常一样愚蠢，但您可以抽象它并向文件添加各种格式，而无需自己搜索。

Answer 2

我试图找到这个问题的解决方案，这里的答案对我不起作用（也许 openpyxl 模块已更新，因此代码已过时）。所以，我做了一些尝试并得出了以下工作代码：

import openpyxl

#Path
wb = openpyxl.load_workbook(r'PathOfTheFile')

#active worksheet data
ws = wb.active    

def wordfinder(searchString):
    for i in range(1, ws.max_row + 1):
        for j in range(1, ws.max_column + 1):
            if searchString == ws.cell(i,j).value:
                print("found")
                print(ws.cell(i,j))          


wordfinder("YourString")

希望这对您有所帮助。 P.S。如果搜索的单词不匹配，则代码不会给出输出

使用 openpyxl 在 Excel sheet 中的文本字符串中搜索单词

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