CODEWITHSUNDEEP
pythonpython-3.xpandas
Martingale
Martingale

Reputation: 523

How to fix 'The truth value of a Series is ambiguous' in Python produced after comparing simple data

after getting data of the Bitcoin (BTC-USD) from Yahoo, I try to create a new column to show if the Close price is higher than the Open price of of each day.

What I want to do is to create a column that shows 1 when the Close is higher than the Open. And a 0 when that condition is not true.

When I try to compare the Close and the Open values I receive the next message : ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

I have tried several things to fix it (like importing the data as a CSV instead directly from Yahoo) but unfortunately I couldn't find the solution.

Below you can see the code that I am using:

import pandas as pd
import pandas_datareader as dr
df = dr.data.get_data_yahoo('btc-usd',start = '01-01-2015', end= '31-12-2018')
df.head(2)

df['X'] = [1 if (df.loc[ei,'Close'] > df.loc[ei,'Open'])  else 0 for ei in df.index] #---> The error is produced in this line of code
df.tail()

Below you can see the error message: 

ValueError                                Traceback (most recent call last)
<ipython-input-45-eb64775bf24f> in <module>
----> 1 df['X'] = [1 if (df.loc[ei,'Close'] > df.loc[ei,'Open'])  else 0 for ei in df.index]
      2 df.tail()

<ipython-input-45-eb64775bf24f> in <listcomp>(.0)
----> 1 df['X'] = [1 if (df.loc[ei,'Close'] > df.loc[ei,'Open'])  else 0 for ei in df.index]
      2 df.tail()

C:\ProgramData\Anaconda3\lib\site-packages\pandas\core\generic.py in __nonzero__(self)
   1476         raise ValueError("The truth value of a {0} is ambiguous. "
   1477                          "Use a.empty, a.bool(), a.item(), a.any() or a.all()."
-> 1478                          .format(self.__class__.__name__))
   1479 
   1480     __bool__ = __nonzero__

ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

I would like to get the result of this formula. Having 1 when the Close is > Open and 0 when the conditional is false.

Upvotes: 1

Views: 688

Answers (2)

Itamar Mushkin
Itamar Mushkin

Reputation: 2905

You can do the series comparison outside of the list comprehension, like this:

df['X'] = 1*(df['Close'] > df['Open'])

It'll work. The series comparison returns a series with True/False values. You can multiply the True by 1 to obtain your desired result.

If you need a more complicated function, you can use an .apply on the df['Close'] > df['Open'] series (I'll show it with simple 1/0 but you can pick it up from there):

df['X'] = (df['Close'] > df['Open']).apply(lambda x: 1 if x else 0)

Upvotes: 0

tawab_shakeel
tawab_shakeel

Reputation: 3739

If you want to place 1 in a column when Close > open just use np.where

import numpy as np
df['X'] = np.where(df['Close']>df['Open'],1,0)

second solution could be

df['X'] = 0
df.loc[df['Close']>df['Open'],"X"] = 1
print(df)

                 High   Low          Open       Close        Volume     Adj Close   X
Date                            
2014-12-31  319.089996  308.890015  311.269989  318.239990  6472822    318.239990   1
2015-01-01  321.359985  313.540009  318.239990  314.890015  4073067    314.890015   0
2015-01-02  316.399994  313.079987  314.890015  315.209991  4673971    315.209991   1
2015-01-03  315.829987  284.890015  315.209991  287.130005  14209564    287.130005  0
2015-01-04  289.940002  255.869995  287.130005  264.720001  24255392    264.720001  0

Upvotes: 1

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