python - 为什么我创建的函数出现 TypeError : f() takes 0 positional arguments but 1 was given

Question

我正在使用 pandas DataFrames，我正在添加新列以进行更高级的分析。我的 f 函数给我一个错误 TypeError: f() takes 0 positional arguments but 1 was given.我不明白为什么，如果您需要知道它的作用，我会在代码中记录我的 f 函数。

from pandas_datareader import data as dreader
import pandas as pd
from datetime import datetime
import dateutil.parser



# Sets the max rows that can be displayed
# when the program is executed
pd.options.display.max_rows = 200



# df is the name of the dataframe, it is 
# reading the csv file containing date loaded
# from yahoo finance(Date,Open,High,Low,Close
# volume,adj close,)the name of the ticker
# is placed before _data.csv i.e. the ticker aapl
# would have a csv file named aapl_data.csv.
df = pd.read_csv("cde_data.csv")



# the following code will allow for filtering of the datafram
# based on the year, day of week (dow), and month. It then gets
# applied to the dataframe and then can be used to sort data i.e
# print(df[(df.year == 2015) & (df.month == 5) & (df.dow == 4)])
# which will give you all the days in the month of May(df.month == 5), 
# that fall on a Thursday(df.dow == 4), in the year 2015 
# (df.year == 2015)
#
#      Month          Day                      Year
# January    = 1  Monday    = 1  The year will be dispaly in a four
# February   = 2  Tuesday   = 2  digit format i.e. 2015
# March      = 3  Wednesday = 3
# April      = 4  Thursday  = 4
# May        = 5  Friday    = 5
# June       = 6
# July       = 7
# August     = 8
# September  = 9
# October    = 10
# November   = 11
# December   = 12

def year(x):
    return(x.year)
def dow(x):
    return(x.isoweekday())
def month(x):
    return(x.month)

# f is a function that checks to see if the up_down column
# has a value that is greater than, less than, or equal to 
# zero. The value in the up_down column is derived from 
# subtracting the opening price of the stock(open column) 
# from closing price of the stock(close column). If up_down 
# has a negative value than the stocks price was Down, a positive
# value then Up, and no change is Flat
def f():
    if up_down > 0:
        x = Up
    elif up_down < 0:
        x = Down
    else:
        x = Flat
return (f)


df.reset_index()

df.Date            = df.Date.apply(dateutil.parser.parse)
df['year']         = df.Date.apply(year)
df['dow']          = df.Date.apply(dow)
df['month']        = df.Date.apply(month)
df['up_down']      = df['Close'] - df['Open']
df['up_down_flat'] = df.up_down.apply(f)


df2= (df[(df.year > 1984) & (df.month == 5) & (df.dow == 1)])
print (df2)

这是错误

Traceback (most recent call last):
  File "dframt1est.py", line 77, in <module>
    df['up_down_flat'] = df.up_down.apply(f)
  File "C:\Users\Zac\AppData\Local\Programs\Python\Python35-32\lib\site-packages\pandas\core\series.py", line 2220, in apply
    mapped = lib.map_infer(values, f, convert=convert_dtype)
  File "pandas\src\inference.pyx", line 1088, in pandas.lib.map_infer (pandas\lib.c:63043)
TypeError: f() takes 0 positional arguments but 1 was given
Press any key to continue . . .

这是一个示例 csv

      Date       Open       High     Low   Close    Volume   Adj Close  \
0    1990-04-12  26.875000  26.875000  26.625  26.625      6100  250.576036
1    1990-04-16  26.500000  26.750000  26.375  26.750       500  251.752449
2    1990-04-17  26.750000  26.875000  26.750  26.875      2300  252.928863
3    1990-04-18  26.875000  26.875000  26.500  26.625      3500  250.576036
4    1990-04-19  26.500000  26.750000  26.500  26.750       700  251.752449
5    1990-04-20  26.750000  26.875000  26.750  26.875      2100  252.928863
6    1990-04-23  26.875000  26.875000  26.750  26.875       700  252.928863
7    1990-04-24  27.000000  27.000000  26.000  26.000      2400  244.693970
8    1990-04-25  25.250000  25.250000  24.875  25.125      9300  236.459076

Answer 1

您正在将 f 作为 apply 中的函数传递。为数据框中的每一行调用该函数，并且需要将该行作为其参数。

另请注意，您返回的是函数本身作为结果；我很确定您打算返回 x 而不是 f。此外，您似乎没有在任何地方定义 Up、Down 和 Flat。