I'm a Pandas newbie, so please bear with me.
Overview: I started with a free-form text file created by a data harvesting script that remotely accessed dozens of different kinds of devices, and multiple instances of each. I used OpenRefine (a truly wonderful tool) to munge that into a CSV that was then input to dataframe df using Pandas in a JupyterLab notebook.
My first inspection of the data showed the 'Timestamp' column was not monotonic. I accessed individual data sources as follows, in this case for the 'T-meter' data source. (The technique was taken from a search result - I don't really understand it, but it worked.)
cond = df['Source']=='T-meter'
rows = df.loc[cond, :]
df_tmeter = pd.DataFrame(columns=df.columns)
df_tmeter = df_tmeter.append(rows, ignore_index=True)
then checked each as follows:
df_tmeter['Timestamp'].is_monotonic
Fortunately, the problem was easy to identify and fix: Some sensors were resetting, then sending bad (but still monotonic) timestamps until their clocks were updated. I wrote the function healing() to cleanly patch such errors, and it worked a treat:
df_tmeter['healed'] = df_tmeter['Timestamp'].apply(healing)
Now for my questions:
How do I get the 'healed' values back into the original df['Timestamp'] column for only the 'T-meter' items in df['Source']?
Given the function healing(), is there a clean way to do this directly on df?
Thanks!
Edit: I first thought I should be using 'views' into df, but other operations on the data would either generate errors, or silently turn the views into copies.
I wrote a wrapper function heal_row() for healing():
def heal_row( row ):
if row['Source'] == 'T-meter': # Redundant check, but safe!
row['Timestamp'] = healing(row['Timestamp'])
return row
then did the following:
df = df.apply(lambda row: row if row['Source'] != 'T-meter' else heal_row(row), axis=1)
This ordering is important, since healing() is stateful based on the prior row(s), and thus can't be the default operation.
So, I'm unclear why doing certain operations on a DF updates it right away, but other times it does not update it unless you re-use the old name or use a new df variable name.
Doesn't this make it really confusing where the last 'real' change is?
First of all, a df behaves like a list in python, meaning that if you make a soft copy of it and change it, the original df changes too. To answer your question you must know that some methods of updating a df, write on a hard copy version of that df (which are indicated by a warning given by pandas letting you know that you are writing on a copy) so the data might not change. the best and most reliable way to change data using pandas is either:
df.at[a_cell] = some_data
or:
df.loc[some_rows, some_columns] = some_data
In the pandas library many times there is an option to change the object inplace such as with the following statement...
df.dropna(axis='index', how='all', inplace=True)
I am curious what is being returned as well as how the object is handled when inplace=True is passed vs. when inplace=False.
Are all operations modifying self when inplace=True? And when inplace=False is a new object created immediately such as new_df = self and then new_df is returned?
If you are trying to close a question where someone should use inplace=True and hasn't, consider replace() method not working on Pandas DataFrame instead.
When inplace=True is passed, the data is renamed in place (it returns nothing), so you'd use:
df.an_operation(inplace=True)
When inplace=False is passed (this is the default value, so isn't necessary), performs the operation and returns a copy of the object, so you'd use:
df = df.an_operation(inplace=False)
In pandas, is inplace = True considered harmful, or not?
TLDR; Yes, yes it is.
inplace, contrary to what the name implies, often does not prevent copies from being created, and (almost) never offers any performance benefits
inplace does not work with method chaining
inplace can lead to SettingWithCopyWarning if used on a DataFrame column, and may prevent the operation from going though, leading to hard-to-debug errors in code
The pain points above are common pitfalls for beginners, so removing this option will simplify the API.
I don't advise setting this parameter as it serves little purpose. See this GitHub issue which proposes the inplace argument be deprecated api-wide.
It is a common misconception that using inplace=True will lead to more efficient or optimized code. In reality, there are absolutely no performance benefits to using inplace=True. Both the in-place and out-of-place versions create a copy of the data anyway, with the in-place version automatically assigning the copy back.
inplace=True is a common pitfall for beginners. For example, it can trigger the SettingWithCopyWarning:
df = pd.DataFrame({'a': [3, 2, 1], 'b': ['x', 'y', 'z']})
df2 = df[df['a'] > 1]
df2['b'].replace({'x': 'abc'}, inplace=True)
# SettingWithCopyWarning:
# A value is trying to be set on a copy of a slice from a DataFrame
Calling a function on a DataFrame column with inplace=True may or may not work. This is especially true when chained indexing is involved.
As if the problems described above aren't enough, inplace=True also hinders method chaining. Contrast the working of
result = df.some_function1().reset_index().some_function2()
As opposed to
temp = df.some_function1()
temp.reset_index(inplace=True)
result = temp.some_function2()
The former lends itself to better code organization and readability.
Another supporting claim is that the API for set_axis was recently changed such that inplace default value was switched from True to False. See GH27600. Great job devs!
The way I use it is
# Have to assign back to dataframe (because it is a new copy)
df = df.some_operation(inplace=False)
Or
# No need to assign back to dataframe (because it is on the same copy)
df.some_operation(inplace=True)
CONCLUSION:
if inplace is False
Assign to a new variable;
else
No need to assign
The inplace parameter:
df.dropna(axis='index', how='all', inplace=True)
in Pandas and in general means:
1. Pandas creates a copy of the original data
2. ... does some computation on it
3. ... assigns the results to the original data.
4. ... deletes the copy.
As you can read in the rest of my answer's further below, we still can have good reason to use this parameter i.e. the inplace operations, but we should avoid it if we can, as it generate more issues, as:
1. Your code will be harder to debug (Actually SettingwithCopyWarning stands for warning you to this possible problem)
2. Conflict with method chaining
So there is even case when we should use it yet?
Definitely yes. If we use pandas or any tool for handeling huge dataset, we can easily face the situation, where some big data can consume our entire memory.
To avoid this unwanted effect we can use some technics like method chaining:
(
wine.rename(columns={"color_intensity": "ci"})
.assign(color_filter=lambda x: np.where((x.hue > 1) & (x.ci > 7), 1, 0))
.query("alcohol > 14 and color_filter == 1")
.sort_values("alcohol", ascending=False)
.reset_index(drop=True)
.loc[:, ["alcohol", "ci", "hue"]]
)
which make our code more compact (though harder to interpret and debug too) and consumes less memory as the chained methods works with the other method's returned values, thus resulting in only one copy of the input data. We can see clearly, that we will have 2 x original data memory consumption after this operations.
Or we can use inplace parameter (though harder to interpret and debug too) our memory consumption will be 2 x original data, but our memory consumption after this operation remains 1 x original data, which if somebody whenever worked with huge datasets exactly knows can be a big benefit.
Final conclusion:
Avoid using inplace parameter unless you don't work with huge data and be aware of its possible issues in case of still using of it.
Save it to the same variable
data["column01"].where(data["column01"]< 5, inplace=True)
Save it to a separate variable
data["column02"] = data["column01"].where(data["column1"]< 5)
But, you can always overwrite the variable
data["column01"] = data["column01"].where(data["column1"]< 5)
FYI: In default inplace = False
When trying to make changes to a Pandas dataframe using a function, we use 'inplace=True' if we want to commit the changes to the dataframe.
Therefore, the first line in the following code changes the name of the first column in 'df' to 'Grades'. We need to call the database if we want to see the resulting database.
df.rename(columns={0: 'Grades'}, inplace=True)
df
We use 'inplace=False' (this is also the default value) when we don't want to commit the changes but just print the resulting database. So, in effect a copy of the original database with the committed changes is printed without altering the original database.
Just to be more clear, the following codes do the same thing:
#Code 1
df.rename(columns={0: 'Grades'}, inplace=True)
#Code 2
df=df.rename(columns={0: 'Grades'}, inplace=False}
Yes, in Pandas we have many functions has the parameter inplace but by default it is assigned to False.
So, when you do df.dropna(axis='index', how='all', inplace=False) it thinks that you do not want to change the orignial DataFrame, therefore it instead creates a new copy for you with the required changes.
But, when you change the inplace parameter to True
Then it is equivalent to explicitly say that I do not want a new copy
of the DataFrame instead do the changes on the given DataFrame
This forces the Python interpreter to not to create a new DataFrame
But you can also avoid using the inplace parameter by reassigning the result to the orignal DataFrame
df = df.dropna(axis='index', how='all')
inplace=True is used depending if you want to make changes to the original df or not.
df.drop_duplicates()
will only make a view of dropped values but not make any changes to df
df.drop_duplicates(inplace = True)
will drop values and make changes to df.
Hope this helps.:)
inplace=True makes the function impure. It changes the original dataframe and returns None. In that case, You breaks the DSL chain.
Because most of dataframe functions return a new dataframe, you can use the DSL conveniently. Like
df.sort_values().rename().to_csv()
Function call with inplace=True returns None and DSL chain is broken. For example
df.sort_values(inplace=True).rename().to_csv()
will throw NoneType object has no attribute 'rename'
Something similar with python’s build-in sort and sorted. lst.sort() returns None and sorted(lst) returns a new list.
Generally, do not use inplace=True unless you have specific reason of doing so. When you have to write reassignment code like df = df.sort_values(), try attaching the function call in the DSL chain, e.g.
df = pd.read_csv().sort_values()...
As Far my experience in pandas I would like to answer.
The 'inplace=True' argument stands for the data frame has to make changes permanent
eg.
df.dropna(axis='index', how='all', inplace=True)
changes the same dataframe (as this pandas find NaN entries in index and drops them).
If we try
df.dropna(axis='index', how='all')
pandas shows the dataframe with changes we make but will not modify the original dataframe 'df'.
If you don't use inplace=True or you use inplace=False you basically get back a copy.
So for instance:
testdf.sort_values(inplace=True, by='volume', ascending=False)
will alter the structure with the data sorted in descending order.
then:
testdf2 = testdf.sort_values( by='volume', ascending=True)
will make testdf2 a copy. the values will all be the same but the sort will be reversed and you will have an independent object.
then given another column, say LongMA and you do:
testdf2.LongMA = testdf2.LongMA -1
the LongMA column in testdf will have the original values and testdf2 will have the decrimented values.
It is important to keep track of the difference as the chain of calculations grows and the copies of dataframes have their own lifecycle.
I'm a veteran of Pandas DataFrame objects, but I'm struggling to find a clean, convenient method for altering the values in a Dask DataFrame column. For a specific example, I'm trying to multiply positive values in a numpy.float column by -1, thereby making them negative. Here is my current method (I'm trying to change the last column in the DataFrame):
cols = df.columns
df[[cols[-1]]] = df[[cols[-1]]]*-1
This seems to work only if the column has a string header, otherwise it adds another column using the index number as a string-type column name for a new column. Is there something akin to the Pandas method of, say, df.iloc[-1,:] = df.iloc[-1,:]*-1 that I can use with a Dask dataframe?
Edit: I'm also trying to implement: df = df.applymap(lambda x: x*-1). This, of course, applies the function to the entire dataframe, but is there a way to apply a function over just one column? Thank you.
first question
If something works for string columns and not for numeric-named columns then that is probably a bug. I recommend raising an issue at https://github.com/dask/dask/issues/new
second question
but is there a way to apply a function over just one column?
You can't apply a single Python function over a dask dataframe that is stored in many pieces directly, however methods like .map_partitions or .reduction may help you to achieve the same result with some cleverness.
in the future we recommend asking separate questions separately on stack overflow
def function_name(df):
for i, row in df.iterrows():
df.set_value(...)
return df
if __name__ == '__main__':
# Assume we have a dataframe called idf
idf = function_name(idf)
In the code above, I pass a dataframe called idf into a function called function_name. In that function, I loop over all rows in the dataframe, make some modifications and return a dataframe which I store back into idf.
I have a feeling that this approach is wasting memory, can someone correct me or point out a better more pythonic approach? Please note that I have a good reason to be using iterrows, even though it makes everything slower, I just want some feedback on the way I am passing dataframe to a function and getting it back
---EDIT--
Based on feedback esp by #marius, here's what I want to know:
By passing dataframe into the function, am I making a new copy of the dataframe? That is the memory wastage I am concerned with