Apply function to dataset to replace dynamically

Question

I have a pandas which is 6x3 and the columns represent times.

I want to replace values on conditions:

def substituteMin(x,n,c,k):
    if x < (1 - c)^n+sqrt(k):
        x = (1 - c)^n+sqrt(k)
    else:
        pass
    return x
df1 = df.apply(lambda x: compareMin(x, x.name))
print (df1)

where c and k are constant. And n is the column name. It gives me the error

"ValueError: ('The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().', u'occurred at index 1.0')"

I tried to look at replace function but they did not work.

> "I have a **panda** which is 6x3 and the columns represent times" - What do you feed it? — ThrowingSpoon, May 11 '17 at 13:23
The error is that you are doing a comparison on a series and returning a single value. Pandas/Python doesn't know what single value to return if part of the series is True and part of the series is False. Hence the "ambiguous" exception. You can use function like all() or any() to reduce the series True/False to one value. — Scott Boston, May 11 '17 at 13:31

score 0 · Answer 1 · edited May 23 '17 at 12:10

0

Here I have simplified your code.

def func(x,n=0,c=1,k=2):
    if x < (1 - c)^n+sqrt(k):
        x = (1 - c)^n+sqrt(k)
    return x
df1 = df.applymap(lambda x: func(x))
print (df1)

Modifications:

applymap to replace element-wise.
Function names were inconsistent: substituteMin, compareMin.
In the function (substituteMin), constant variables (n,c,k) can be set in the function only.

(There is plenty to comment and edit in your question. Currently, I don't have privileges.)

edited May 23 '17 at 12:10

Community

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answered May 11 '17 at 14:15

Thanks but how can I pass n to be the column name? I would do "lambda x: func(x, x.name)" but it does not work – Thegamer23 May 15 '17 at 07:25
that's because `x` would not have have an attribute `name`. to accomplish your objective I guess you should simply iterate over rows and columns (nested). eg. `for col in df: for row in df.index: y=func(df.loc[row,col],col)` Note: go to new line after `:`s. it's not a one liner like `lambda`. – May 15 '17 at 13:58

Apply function to dataset to replace dynamically

1 Answers1