Pandas: Selecting column from data frame - python

Pandas beginner here. I'm looking to return a full column's data and I've seen a couple of different methods for this.
What is the difference between the two entries below, if any? It looks like they return the same thing.
loansData['int_rate']
loansData.int_rate

The latter is basically syntactic sugar for the former. There are (at least) a couple of gotchas:
If the name of the column is not a valid Python identifier (e.g., if the column name is my column name?!, you must use the former.
Somewhat surprisingly, you can only use the former form to completely correctly add a new column (see, e.g., here).
Example for latter statement:
import pandas as pd
df = pd.DataFrame({'a': range(4)})
df.b = range(4)
>> df.columns
Index([u'a'], dtype='object')
For some reason, though, df.b returns the correct results.

They do return the same thing. The column names in pandas are akin to dictionary keys that refer to a series. The column names themselves are named attributes that are part of the dataframe object.
The first method is preferred as it allows for spaces and other illegal operators.
For a more complete explanation, I recommend you take a look at this article:
http://byumcl.bitbucket.org/bootcamp2013/labs/pd_types.html#pandas-types
Search 'Access using dict notation' to find the examples where they show that these two methods return identical values.

They're the same but for me the first method handles spaces in column names and illegal characters so is preferred, example:
In [115]:
df = pd.DataFrame(columns=['a', ' a', '1a'])
df
Out[115]:
Empty DataFrame
Columns: [a, a, 1a]
Index: []
In [116]:
print(df.a) # works
print([' a']) # works
print(df.1a) # error
File "<ipython-input-116-4fa4129a400e>", line 3
print(df.1a)
^
SyntaxError: invalid syntax
Really when you use dot . it's trying to find a key as an attribute, if for some reason you have used column names that match an attribute then using dot will not do what you expect.
Example:
In [121]:
df = pd.DataFrame(columns=['index'], data = np.random.randn(3))
df
Out[121]:
index
0 0.062698
1 -1.066654
2 -1.560549
In [122]:
df.index
Out[122]:
Int64Index([0, 1, 2], dtype='int64')
The above has now shown the index as opposed to the column 'index'

In case if you are working on any ML projects and you want to extract feature and target variables separately and need to have them separably.
Below code will be useful: This is selecting features through indexing as a list and applying them to the dataframe. in this code data is DF.
len_col=len(data.columns)
total_col=list(data.columns)
Target_col_Y=total_col[-1]
Feature_col_X=total_col[0:-1]
print('The dependent variable is')
print(Target_col_Y)
print('The independent variables are')
print(Feature_col_X)
The output for the same can be obtained as given below:
The dependent variable is
output
The independent variables are
['age', 'job', 'marital', 'education','day_of_week', ... etc]

Related

Getting not in index error in Pandas Dataframe [duplicate]

I'm struggling to understand the concept behind column naming conventions, given that one of the following attempts to create a new column appears to fail:
from numpy.random import randn
import pandas as pd
df = pd.DataFrame({'a':range(0,10,2), 'c':range(0,1000,200)},
columns=list('ac'))
df['b'] = 10*df.a
df
gives the following result:
Yet, if I were to try to create column b by substituting with the following line, there is no error message, yet the dataframe df remains with only the columns a and c.
df.b = 10*df.a ### rather than the previous df['b'] = 10*df.a ###
What has pandas done and why is my command incorrect?
What you did was add an attribute b to your df:
In [70]:
df.b = 10*df.a
df.b
Out[70]:
0 0
1 20
2 40
3 60
4 80
Name: a, dtype: int32
but we see that no new column has been added:
In [73]:
df.columns
Out[73]:
Index(['a', 'c'], dtype='object')
which means we get a KeyError if we tried df['b'], to avoid this ambiguity you should always use square brackets when assigning.
for instance if you had a column named index or sum or max then doing df.index would return the index and not the index column, and similarly df.sum and df.max would screw up those df methods.
I strongly advise to always use square brackets, it avoids any ambiguity and the latest ipython is able to resolve column names using square brackets. It's also useful to think of a dataframe as a dict of series in which it makes sense to use square brackets for assigning and returning a column
Always use square brackets for assigning columns
Dot notation is a convenience for accessing columns in a dataframe. If they conflict with existing properties (e.g. if you had a column named 'max'), then you need to use square brackets to access that column, e.g. df['max']. You also need to use square brackets when the column name contains spaces, e.g. df['max value'].
A DataFrame is just an object which has the usual properties and methods. If you use dot notation for assignment, you are creating a property or method for the dataframe object. So df.val = 2 will assign df with a property val that has a value of two. This is very different from df['val'] = 2 which creates a new column in the dataframe and assigns each element in that column the value of two.
To be safe, using square bracket notation will always provide the correct result.
As an aside, your columns=list('ac')) doesn't do anything, as you are just creating a variable named columns that is never used. You may have meant df.columns = list('ac'), but you already assigned those in the creation of the dataframe, so I'm not sure what the intent is with this line of code. And remember that dictionaries are unordered, so that pd.DataFrame({'a': [...], 'b': [...]}) could potentially return a dataframe with columns ['b', 'a']. If this were the case, then assigning column names could potentially mix up the column headers.
The issue has to do with how properties are handled in python. There is no restriction in python of setting a new properties for a class, so for example you could do something like
df.myspecialstuff = ["dog", "cat", 5]
So when you do assignment like
df.b = 10*df.a
It is ambiguous whether you want to add a property or a new column, and a property is set. The easiest way to actually see what is going on with this is to use pdb and step through the code
import pdb
x = df.a
pdb.run("df.a1 = x")
This will step into the __setattr__() whereas pdb.run("df['a2'] = x") will step into __setitem__()

Does iloc use the indices or the place of the row

I have extracted few rows from a dataframe to a new dataframe. In this new dataframe old indices remain. However, when i want to specify range from this new dataframe i used it like new indices, starting from zero. Why did it work? Whenever I try to use the old indices it gives an error.
germany_cases = virus_df_2[virus_df_2['location'] == 'Germany']
germany_cases = germany_cases.iloc[:190]
This is the code. The rows that I extracted from the dataframe virus_df_2 have indices between 16100 and 16590. I wanted to take the first 190 rows. in the second line of code i used iloc[:190] and it worked. However, when i tried to use iloc[16100:16290] it gave an error. What could be the reason?
In pandas there are two attributes, loc and iloc.
The iloc is, as you have noticed, an indexing based on the order of the rows in memory, so there you can reference the nth line using iloc[n].
In order to reference rows using the pandas indexing, which can be manually altered and can not only be integers but also strings or other objects that are hashable (have the __hash__ method defined), you should use loc attribute.
In your case, iloc raises an error because you are trying to access a range that is outside the region defined by your dataframe. You can try loc instead and it will be ok.
At first it will be hard to grasp the indexing notation, but it can be very helpful in some circumstances, like for example sorting or performing grouping operations.
Quick example that might help:
df = pd.DataFrame(
dict(
France=[1, 2, 3],
Germany=[4, 5, 6],
UK=['x', 'y', 'z'],
))
df = df.loc[:,"Germany"].iloc[1:2]
Out:
1 5
Name: Germany, dtype: int64
Hope I could help.

Filters in Pandas - why doesn't this work?

This is a basic question so apologies in advance.
I am using Pandas and I am grouping data with the following line:
page_serp_df.groupby([page_serp_df.meta_keywords_1_length]).count()['keyword']
This is referencing the following:
The data frame: [page_serp_df]
Grouping by the column: meta_keywords_1_length
Counting with the filter: keyword column
What I don't understand is why does the filtering condition have to be ['keyword'] i.e. a string in quotes?
For example, this doesn't work and it is very counterintuituve to me:
page_serp_df.groupby([page_serp_df.meta_keywords_1_length]).count()[page_serp_df.keyword]
Thanks in advance!
I think there is a misunderstanding on what the .count() method returns.
Try to follow this example:
Create a sample data frame
df = pd.DataFrame({
'A':[0,1,0,1, 1],
'B':[100,200,300, 400, 500],
'C': [1,2,3,4,5]
})
This is what the count() method will return after groupby
# similarly to your example I am grouping by A and counting
df.groupby([df.A]).count()
As you can see, the count() method returns a dataframe itself, having the count of each other column values for the column where the grouped column has the same value.
After that, you can query for a specific column form the return of count() like this
df.groupby([df.A]).count()['C']
But the second case in your example, which in my example would correspond to df.groupby([df.A]).count()[df.C]
Will throw an error!
In fact, you would query a dataframe (in this case df.groupby([df.A]).count()) via a pandas Series but as you know you need a string or a column from df.columns.
You can check yourself that df.C and 'C' are two very different variable types.
print(type(df.C))
print(type('C'))
# <class 'pandas.core.series.Series'>
# <class 'str'>
If for some reason your code still works with the equivalent of df.C there might be some contingency like the only value of the df.C is a string with the same name of a column.. or something unintentional like that.

Why is 'series' type not a special case of 'dataframe' type in pandas? [duplicate]

In Pandas, when I select a label that only has one entry in the index I get back a Series, but when I select an entry that has more then one entry I get back a data frame.
Why is that? Is there a way to ensure I always get back a data frame?
In [1]: import pandas as pd
In [2]: df = pd.DataFrame(data=range(5), index=[1, 2, 3, 3, 3])
In [3]: type(df.loc[3])
Out[3]: pandas.core.frame.DataFrame
In [4]: type(df.loc[1])
Out[4]: pandas.core.series.Series
Granted that the behavior is inconsistent, but I think it's easy to imagine cases where this is convenient. Anyway, to get a DataFrame every time, just pass a list to loc. There are other ways, but in my opinion this is the cleanest.
In [2]: type(df.loc[[3]])
Out[2]: pandas.core.frame.DataFrame
In [3]: type(df.loc[[1]])
Out[3]: pandas.core.frame.DataFrame
The TLDR
When using loc
df.loc[:] = Dataframe
df.loc[int] = Dataframe if you have more than one column and Series if you have only 1 column in the dataframe
df.loc[:, ["col_name"]] = Dataframe if you have more than one row and Series if you have only 1 row in the selection
df.loc[:, "col_name"] = Series
Not using loc
df["col_name"] = Series
df[["col_name"]] = Dataframe
You have an index with three index items 3. For this reason df.loc[3] will return a dataframe.
The reason is that you don't specify the column. So df.loc[3] selects three items of all columns (which is column 0), while df.loc[3,0] will return a Series. E.g. df.loc[1:2] also returns a dataframe, because you slice the rows.
Selecting a single row (as df.loc[1]) returns a Series with the column names as the index.
If you want to be sure to always have a DataFrame, you can slice like df.loc[1:1]. Another option is boolean indexing (df.loc[df.index==1]) or the take method (df.take([0]), but this used location not labels!).
Use df['columnName'] to get a Series and df[['columnName']] to get a Dataframe.
You wrote in a comment to joris' answer:
"I don't understand the design
decision for single rows to get converted into a series - why not a
data frame with one row?"
A single row isn't converted in a Series.
It IS a Series: No, I don't think so, in fact; see the edit
The best way to think about the pandas data structures is as flexible
containers for lower dimensional data. For example, DataFrame is a
container for Series, and Panel is a container for DataFrame objects.
We would like to be able to insert and remove objects from these
containers in a dictionary-like fashion.
http://pandas.pydata.org/pandas-docs/stable/overview.html#why-more-than-1-data-structure
The data model of Pandas objects has been choosen like that. The reason certainly lies in the fact that it ensures some advantages I don't know (I don't fully understand the last sentence of the citation, maybe it's the reason)
.
Edit : I don't agree with me
A DataFrame can't be composed of elements that would be Series, because the following code gives the same type "Series" as well for a row as for a column:
import pandas as pd
df = pd.DataFrame(data=[11,12,13], index=[2, 3, 3])
print '-------- df -------------'
print df
print '\n------- df.loc[2] --------'
print df.loc[2]
print 'type(df.loc[1]) : ',type(df.loc[2])
print '\n--------- df[0] ----------'
print df[0]
print 'type(df[0]) : ',type(df[0])
result
-------- df -------------
0
2 11
3 12
3 13
------- df.loc[2] --------
0 11
Name: 2, dtype: int64
type(df.loc[1]) : <class 'pandas.core.series.Series'>
--------- df[0] ----------
2 11
3 12
3 13
Name: 0, dtype: int64
type(df[0]) : <class 'pandas.core.series.Series'>
So, there is no sense to pretend that a DataFrame is composed of Series because what would these said Series be supposed to be : columns or rows ? Stupid question and vision.
.
Then what is a DataFrame ?
In the previous version of this answer, I asked this question, trying to find the answer to the Why is that? part of the question of the OP and the similar interrogation single rows to get converted into a series - why not a data frame with one row? in one of his comment,
while the Is there a way to ensure I always get back a data frame? part has been answered by Dan Allan.
Then, as the Pandas' docs cited above says that the pandas' data structures are best seen as containers of lower dimensional data, it seemed to me that the understanding of the why would be found in the characteristcs of the nature of DataFrame structures.
However, I realized that this cited advice must not be taken as a precise description of the nature of Pandas' data structures.
This advice doesn't mean that a DataFrame is a container of Series.
It expresses that the mental representation of a DataFrame as a container of Series (either rows or columns according the option considered at one moment of a reasoning) is a good way to consider DataFrames, even if it isn't strictly the case in reality. "Good" meaning that this vision enables to use DataFrames with efficiency. That's all.
.
Then what is a DataFrame object ?
The DataFrame class produces instances that have a particular structure originated in the NDFrame base class, itself derived from the PandasContainer base class that is also a parent class of the Series class.
Note that this is correct for Pandas until version 0.12. In the upcoming version 0.13, Series will derive also from NDFrame class only.
# with pandas 0.12
from pandas import Series
print 'Series :\n',Series
print 'Series.__bases__ :\n',Series.__bases__
from pandas import DataFrame
print '\nDataFrame :\n',DataFrame
print 'DataFrame.__bases__ :\n',DataFrame.__bases__
print '\n-------------------'
from pandas.core.generic import NDFrame
print '\nNDFrame.__bases__ :\n',NDFrame.__bases__
from pandas.core.generic import PandasContainer
print '\nPandasContainer.__bases__ :\n',PandasContainer.__bases__
from pandas.core.base import PandasObject
print '\nPandasObject.__bases__ :\n',PandasObject.__bases__
from pandas.core.base import StringMixin
print '\nStringMixin.__bases__ :\n',StringMixin.__bases__
result
Series :
<class 'pandas.core.series.Series'>
Series.__bases__ :
(<class 'pandas.core.generic.PandasContainer'>, <type 'numpy.ndarray'>)
DataFrame :
<class 'pandas.core.frame.DataFrame'>
DataFrame.__bases__ :
(<class 'pandas.core.generic.NDFrame'>,)
-------------------
NDFrame.__bases__ :
(<class 'pandas.core.generic.PandasContainer'>,)
PandasContainer.__bases__ :
(<class 'pandas.core.base.PandasObject'>,)
PandasObject.__bases__ :
(<class 'pandas.core.base.StringMixin'>,)
StringMixin.__bases__ :
(<type 'object'>,)
So my understanding is now that a DataFrame instance has certain methods that have been crafted in order to control the way data are extracted from rows and columns.
The ways these extracting methods work are described in this page:
http://pandas.pydata.org/pandas-docs/stable/indexing.html#indexing
We find in it the method given by Dan Allan and other methods.
Why these extracting methods have been crafted as they were ?
That's certainly because they have been appraised as the ones giving the better possibilities and ease in data analysis.
It's precisely what is expressed in this sentence:
The best way to think about the pandas data structures is as flexible
containers for lower dimensional data.
The why of the extraction of data from a DataFRame instance doesn't lies in its structure, it lies in the why of this structure. I guess that the structure and functionning of the Pandas' data structure have been chiseled in order to be as much intellectually intuitive as possible, and that to understand the details, one must read the blog of Wes McKinney.
If the objective is to get a subset of the data set using the index, it is best to avoid using loc or iloc. Instead you should use syntax similar to this :
df = pd.DataFrame(data=range(5), index=[1, 2, 3, 3, 3])
result = df[df.index == 3]
isinstance(result, pd.DataFrame) # True
result = df[df.index == 1]
isinstance(result, pd.DataFrame) # True
every time we put [['column name']] it returns Pandas DataFrame object,
if we put ['column name'] we got Pandas Series object
If you also select on the index of the dataframe then the result can be either a DataFrame or a Series or it can be a Series or a scalar (single value).
This function ensures that you always get a list from your selection (if the df, index and column are valid):
def get_list_from_df_column(df, index, column):
df_or_series = df.loc[index,[column]]
# df.loc[index,column] is also possible and returns a series or a scalar
if isinstance(df_or_series, pd.Series):
resulting_list = df_or_series.tolist() #get list from series
else:
resulting_list = df_or_series[column].tolist()
# use the column key to get a series from the dataframe
return(resulting_list)

Pandas DataFrame naming only 1 column

Is there a way with Pandas Dataframe to name only the first or first and second column even if there's 4 columns :
Here
for x in range(1, len(table2_query) + 1):
if x == 1:
cursor.execute(table2_query[x])
df = pd.DataFrame(data=cursor.fetchall(), columns=['Q', col_name[x-1]])
and it gives me this :
AssertionError: 2 columns passed, passed data had 4 columns
Consider the df:
df = pd.DataFrame(np.arange(8).reshape(2, 4), columns=list('ABCD'))
df
then use rename and pass a dictionary with the name changes to the argument columns:
df.rename(columns=dict(A='a', B='b'))
Instantiating a DataFrame while only naming a subset of the columns
When constructing a dataframe with pd.DataFrame, you either don't pass an index/columns argument and let pandas auto-generate the index/columns object, or you pass one in yourself. If you pass it in yourself, it must match the dimensions of your data. The trouble of mimicking the auto-generation of pandas while augmenting just the ones you want is not worth the trouble and is ugly and is probably non-performant. In other words, I can't even think of a good reason to do it.
On the other hand, it is super easy to rename the columns/index values. In fact, we can rename just a few. I think below is more in line with the spirit of your question:
df = pd.DataFrame(np.arange(8).reshape(2, 4)).rename(columns=str).rename(columns={'1': 'A', '3': 'F'})
df

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