通过与以前的标题名称连接来重命名 pandas 中的列标题

Question

我有一组列名称，其中名称描述多次存在。

['Sl.No', 'Job', 'Description', 'Vendor', 'Description', 'WO No',

       'Accounting Center ', 'Description', 'Nature Of Work', 'WO Type',

       'WO Date', 'WO From Date', 'WO To Date', 'Bill No', 'Running Bill No',

       'Bill Date', 'Bill Status', 'Voucher No', 'Voucher Date',

       'Bank \nVoucher No', 'Bank Voucher\n Date', ' Paid Amount', 'Currency',

       'WO Amt', 'Bill Amt', 'Service Tax Amt', 'VAT Amt', 'Total Tax \nAmt',

       ' Advance Amt', 'Gross Amt', 'Deduction Amt', 'Net Amt']

我希望名称“Description”与其后面的任何标头名称连接起来

例如

“Job”、“Description”将变为“Job_Description”

“Vendor”、“Description”将变为“Vendor_Description”等

我需要一个不引入任何硬编码的逻辑

['Sl.No', 'Job', 'Job_Description', 'Vendor', 'Vendor_Description',....]

Answer 1

你需要找出哪里Description存在using pd.Series.eq，然后使用pd.Index.to_series，这样我们才能使用pd.Series.shift。cols = df.columns.to_series()m = cols.eq('Description') #would mark True where value is `Description`out = cols.shift().str.strip().str.replace('\s+','_') + '_' + colsout[~m] = colsout.to_list()['Sl.No', 'Job', 'Job_Description', 'Vendor', 'Vendor_Description', 'WO No', 'Accounting Center ', 'Accounting_Center_Description', 'Nature Of Work', 'WO Type', 'WO Date', 'WO From Date', 'WO To Date', 'Bill No', 'Running Bill No', 'Bill Date', 'Bill Status', 'Voucher No', 'Voucher Date', 'Bank \nVoucher No', 'Bank Voucher\n Date', ' Paid Amount', 'Currency', 'WO Amt', 'Bill Amt', 'Service Tax Amt', 'VAT Amt', 'Total Tax \nAmt', ' Advance Amt', 'Gross Amt', 'Deduction Amt', 'Net Amt']

Answer 2

让cols成为您原始列表的名称。那么代码可能是：D = "Description"previous_current = zip([None] + cols[:-1], cols)df.columns = [curr if curr != D else prev + "_" + D  for prev, curr in previous_current]结果：>>> df.columnsIndex(['Sl.No', 'Job', 'Job_Description', 'Vendor', 'Vendor_Description',       'WO No', 'Accounting Center ', 'Accounting Center _Description',       'Nature Of Work', 'WO Type', 'WO Date', 'WO From Date', 'WO To Date',       'Bill No', 'Running Bill No', 'Bill Date', 'Bill Status', 'Voucher No',       'Voucher Date', 'Bank \nVoucher No', 'Bank Voucher\n Date',       ' Paid Amount', 'Currency', 'WO Amt', 'Bill Amt', 'Service Tax Amt',       'VAT Amt', 'Total Tax \nAmt', ' Advance Amt', 'Gross Amt',       'Deduction Amt', 'Net Amt'],      dtype='object')说明：我们构建了一些作为对列表的东西(previous, current)：>>> previous_current = zip([None] + cols[:-1], cols)>>> list(previous_current)[(None, 'Sl.No'), ('Sl.No', 'Job'), ('Job', 'Description'), ('Description', 'Vendor'), ('Vendor', 'Description'), ('Description', 'WO No'), ('WO No', 'Accounting Center '), ('Accounting Center ', 'Description'), ('Description', 'Nature Of Work'), ('Nature Of Work', 'WO Type'), ('WO Type', 'WO Date'), ('WO Date', 'WO From Date'), ('WO From Date', 'WO To Date'), ('WO To Date', 'Bill No'), ('Bill No', 'Running Bill No'), ('Running Bill No', 'Bill Date'), ('Bill Date', 'Bill Status'), ('Bill Status', 'Voucher No'), ('Voucher No', 'Voucher Date'), ('Voucher Date', 'Bank \nVoucher No'), ('Bank \nVoucher No', 'Bank Voucher\n Date'), ('Bank Voucher\n Date', ' Paid Amount'), (' Paid Amount', 'Currency'), ('Currency', 'WO Amt'), ('WO Amt', 'Bill Amt'), ('Bill Amt', 'Service Tax Amt'), ('Service Tax Amt', 'VAT Amt'), ('VAT Amt', 'Total Tax \nAmt'), ('Total Tax \nAmt', ' Advance Amt'), (' Advance Amt', 'Gross Amt'), ('Gross Amt', 'Deduction Amt'), ('Deduction Amt', 'Net Amt')]然后我们迭代它，并通过该对的第二个元素（当前列标签）我们决定是否保持原样，或者将其附加到前一列的标签（该对的第一个元素）。

Answer 3

让cols成为您原始列表的名称。那么代码可能是：D = "Description"previous_current = zip([None] + cols[:-1], cols)df.columns = [curr if curr != D else prev + "_" + D  for prev, curr in previous_current]结果：>>> df.columnsIndex(['Sl.No', 'Job', 'Job_Description', 'Vendor', 'Vendor_Description',       'WO No', 'Accounting Center ', 'Accounting Center _Description',       'Nature Of Work', 'WO Type', 'WO Date', 'WO From Date', 'WO To Date',       'Bill No', 'Running Bill No', 'Bill Date', 'Bill Status', 'Voucher No',       'Voucher Date', 'Bank \nVoucher No', 'Bank Voucher\n Date',       ' Paid Amount', 'Currency', 'WO Amt', 'Bill Amt', 'Service Tax Amt',       'VAT Amt', 'Total Tax \nAmt', ' Advance Amt', 'Gross Amt',       'Deduction Amt', 'Net Amt'],      dtype='object')说明：我们构建了一些作为对列表的东西(previous, current)：>>> previous_current = zip([None] + cols[:-1], cols)>>> list(previous_current)[(None, 'Sl.No'), ('Sl.No', 'Job'), ('Job', 'Description'), ('Description', 'Vendor'), ('Vendor', 'Description'), ('Description', 'WO No'), ('WO No', 'Accounting Center '), ('Accounting Center ', 'Description'), ('Description', 'Nature Of Work'), ('Nature Of Work', 'WO Type'), ('WO Type', 'WO Date'), ('WO Date', 'WO From Date'), ('WO From Date', 'WO To Date'), ('WO To Date', 'Bill No'), ('Bill No', 'Running Bill No'), ('Running Bill No', 'Bill Date'), ('Bill Date', 'Bill Status'), ('Bill Status', 'Voucher No'), ('Voucher No', 'Voucher Date'), ('Voucher Date', 'Bank \nVoucher No'), ('Bank \nVoucher No', 'Bank Voucher\n Date'), ('Bank Voucher\n Date', ' Paid Amount'), (' Paid Amount', 'Currency'), ('Currency', 'WO Amt'), ('WO Amt', 'Bill Amt'), ('Bill Amt', 'Service Tax Amt'), ('Service Tax Amt', 'VAT Amt'), ('VAT Amt', 'Total Tax \nAmt'), ('Total Tax \nAmt', ' Advance Amt'), (' Advance Amt', 'Gross Amt'), ('Gross Amt', 'Deduction Amt'), ('Deduction Amt', 'Net Amt')]