Search a word in a text string in Excel sheet using openpyxl
Question:
I’m trying to search for a word in a cell that has a text string that looks like this (Energy;Green Buildings;High Performance Buildings). Here is the code I wrote, I get a syntax error
for row in ws.iter_rows('D2:D11'):
for cell in row:
if 'Energy' in ws.cell.value :
Print 'yes'
Obviously, I don’t want to print yes, this was to test the search function.
Additionally, I want to get the cell location, and then tell openpyxl to assign a color to a cell in the same row under column E. here is a snap shot of my Excel sheet.
I know how to assign a color using this command
c.fill = PatternFill(start_color='FFFFE0', end_color='FFFFE0'
fill_type='solid')
I just need help getting the cell location (the cell that has a matching text) and assign its row number to another cell in column E
UPDATE: I wrote this code below that is working fine for me:
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")
Answers:
For an approach to searching I suggest you look at the 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. Used this myself for the first time last week. The Excel API is stupid as usual but you could abstract it and add all kinds of formatting to the file without ever searching yourself.
I was trying to find a solution for this question, the answers here didn’t work for me (maybe the openpyxl module was updated and hence the code has become obsolete). So, I tried a little and came up with the following working code:
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")
Hope this helps.
P.S. The code doesnt give an output if there in no match for the word searched
I think this will help you…
from openpyxl import load_workbook
wb = load_workbook('C:\UsersluxmanDesktopT11.xlsx')
ws=wb.active
for row in ws.iter_rows():
for cell in row:
if cell.value == "Status":
#print(ws.cell(row=cell.row, column=3).value)
print('yes')
I’m trying to search for a word in a cell that has a text string that looks like this (Energy;Green Buildings;High Performance Buildings). Here is the code I wrote, I get a syntax error
for row in ws.iter_rows('D2:D11'):
for cell in row:
if 'Energy' in ws.cell.value :
Print 'yes'
Obviously, I don’t want to print yes, this was to test the search function.
Additionally, I want to get the cell location, and then tell openpyxl to assign a color to a cell in the same row under column E. here is a snap shot of my Excel sheet.
I know how to assign a color using this command
c.fill = PatternFill(start_color='FFFFE0', end_color='FFFFE0')
fill_type='solid'
I just need help getting the cell location (the cell that has a matching text) and assign its row number to another cell in column E
UPDATE: I wrote this code below that is working fine for me:
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")
For an approach to searching I suggest you look at the 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. Used this myself for the first time last week. The Excel API is stupid as usual but you could abstract it and add all kinds of formatting to the file without ever searching yourself.
I was trying to find a solution for this question, the answers here didn’t work for me (maybe the openpyxl module was updated and hence the code has become obsolete). So, I tried a little and came up with the following working code:
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")
Hope this helps.
P.S. The code doesnt give an output if there in no match for the word searched
I think this will help you…
from openpyxl import load_workbook
wb = load_workbook('C:\UsersluxmanDesktopT11.xlsx')
ws=wb.active
for row in ws.iter_rows():
for cell in row:
if cell.value == "Status":
#print(ws.cell(row=cell.row, column=3).value)
print('yes')