Pandas：如何改变列的所有值？ [英] Pandas: how to change all the values of a column?

问题描述

我有一个包含Date的列的数据框，并且希望此列中的所有值具有相同的值（仅限年份）。示例：

I have a data frame with a column called "Date" and want all the values from this column to have the same value (the year only). Example:

City     Date
Paris    01/04/2004
Lisbon   01/09/2004
Madrid   2004
Pekin    31/2004

我想要的是：

City     Date
Paris    2004
Lisbon   2004
Madrid   2004
Pekin    2004

以下是我的代码：

fr61_70xls = pd.ExcelFile('AMADEUS FRANCE 1961-1970.xlsx')

#Here we import the individual sheets and clean the sheets    
years=(['1961','1962','1963','1964','1965','1966','1967','1968','1969','1970'])

fr={}

header=(['City','Country','NACE','Cons','Last_year','Op_Rev_EUR_Last_avail_yr','BvD_Indep_Indic','GUO_Name','Legal_status','Date_of_incorporation','Legal_status_date'])

for year in years:
    # save every sheet in variable fr['1961'], fr['1962'] and so on
    fr[year]=fr61_70xls.parse(year,header=0,parse_cols=10)
    fr[year].columns=header
    # drop the entire Legal status date column
    fr[year]=fr[year].drop(['Legal_status_date','Date_of_incorporation'],axis=1)
    # drop every row where GUO Name is empty
    fr[year]=fr[year].dropna(axis=0,how='all',subset=[['GUO_Name']])
    fr[year]=fr[year].set_index(['GUO_Name','Date_of_incorporation'])

这发生在我的DataFrames中，例如 fr ['1961 '] 的值 Date_of_incorporation 可以是任何（字符串，整数等），所以也许最好完全擦除此列然后将另一个只有年份的列附加到DataFrames？

It happens that in my DataFrames, called for example fr['1961'] the values of Date_of_incorporation can be anything (strings, integer, and so on), so maybe it would be best to completely erase this column and then attach another column with only the year to the DataFrames?

推荐答案

请使用向量化字符串方法更直接地执行此操作：

As @DSM points out, you can do this more directly using the vectorised string methods:

df['Date'].str[-4:].astype(int)

或使用extract（假设每个字符串中某处只有一组长度为4的数字）：

Or using extract (assuming there is only one set of digits of length 4 somewhere in each string):

df['Date'].str.extract('(?P<year>\d{4})').astype(int)

另一种略微更灵活的方法，可能是使用 apply （或等效地 map ）：

An alternative slightly more flexible way, might be to use apply (or equivalently map) to do this:

df['Date'] = df['Date'].apply(lambda x: int(str(x)[-4:]))
             #  converts the last 4 characters of the string to an integer

lambda函数输入 Date 并将其转换为一年。

您可以（也许应该）将此更详细地写为：

The lambda function, is taking the input from the Date and converting it to a year.
You could (and perhaps should) write this more verbosely as:

def convert_to_year(date_in_some_format);
    date_as_string = str(date_in_some_format)
    year_as_string = date_in_some_format[-4:] # last four characters
    return int(year_as_string)

df['Date'] = df['Date'].apply(convert_to_year)

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