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Answer : A helper function for appending DataFrame to existing Excel file: def append_df_to_excel(filename, df, sheet_name='Sheet1', startrow=None, truncate_sheet=False, **to_excel_kwargs): """ Append a DataFrame [df] to existing Excel file [filename] into [sheet_name] Sheet. If [filename] doesn't exist, then this function will create it. Parameters: filename : File path or existing ExcelWriter (Example: '/path/to/file.xlsx') df : dataframe to save to workbook sheet_name : Name of sheet which will contain DataFrame. (default: 'Sheet1') startrow : upper left cell row to dump data frame. Per default (startrow=None) calculate the last row in the existing DF and write to the next row... truncate_sheet : truncate (remove and recreate) [sheet_name] before ...

Answer : Write a proper comprehension, without append. >>> [i for i in range(10)] [0, 1, 2, 3, 4, 5, 6, 7, 8, 9] >>> list(i for i in range(10)) [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

Answer : I think you can use concat : print pd.concat([t1, t2, t3, t4, t5]) Maybe you can ignore_index : print pd.concat([t1, t2, t3, t4, t5], ignore_index=True) More info in docs. Have you simply tried using a list as argument of append? Or am I missing anything? import numpy as np import pandas as pd dates = np.asarray(pd.date_range('1/1/2000', periods=8)) df1 = pd.DataFrame(np.random.randn(8, 4), index=dates, columns=['A', 'B', 'C', 'D']) df2 = df1.copy() df3 = df1.copy() df = df1.append([df2, df3]) print df

Answer : You could use .before() to add a sibling before the element: $("#wrapper .content:last").before('<div class="content"><div class="subcontent">Third</div></div>'); .insertBefore() does the same thing with a different syntax, namely that you select the element to be added, and pass the element you want to add it before. $("#wrapper .content:last").before('<div class="content"><div class="subcontent">Third</div></div>'); <script src="https://ajax.googleapis.com/ajax/libs/jquery/1.11.1/jquery.min.js"></script> <div id="wrapper"> <div class="content"> <div class="subcontent"> First </div> </div> <div class="content"> <div class="subcontent"> Second </div> ...

Answer : If you're interested in creating a new dict without using intermediary storage: (this is faster, and in my opinion, cleaner than using dict.items()) dic2 = dict(dic0, **dic1) Or if you're happy to use one of the existing dicts: dic0.update(dic1) Here are quite a few ways to add dictionaries. You can use Python3's dictionary unpacking feature. ndic = {**dic0, **dic1} Or create a new dict by adding both items. ndic = dict(dic0.items() + dic1.items()) If your ok to modify dic0 dic0.update(dic1) If your NOT ok to modify dic0 ndic = dic0.copy() ndic.update(dic1) If all the keys in one dict are ensured to be strings ( dic1 in this case, of course args can be swapped) ndic = dict(dic0, **dic1) In some cases it may be handy to use dict comprehensions (Python 2.7 or newer), Especially if you want to filter out or transform some keys/values at the same time. ndic = {k: v for d in (dic0, dic1) for k, v in d.items()} >>> dic0 = {...