numpy get rows where value in list within column - python

I have kind of an odd numpy ndarray, it looks like this:
[[0,1,3,list([0,1])],
[0,0,0,list([0])],
[0,0,0,list([])],
[1,1,1,list([1,2,3,4,5,6,7])]]
so, the column at index 3 is actually of type list.
I would like to extract the rows which contain a key value in the list contained in index 3, so for instance, something like magic_function(matrix, 0) might return [[0,1,3,list([0,1])], [0,0,0,list([0])]], as the rows at index 0 and 1 contain the value 0 in the list at index 3.
I tried using a combination of np.where and np.isin, but couldn't quite get it working in a way that was elegant (things like matrix[np.where(np.isin(matrix[:,3], 0))]). I would prefer the fastest approach, which I beleive would be an approach using numpy rather than iteration in python.

Related

Selecting by index -1 in a df column / time series throws error

Let's assume we have a simple dataframe like this:
df = pd.DataFrame({'col1':[1,2,3], 'col2':[10,20,30]})
Then I can select elements like this
df.col2[0] or df.col2[1]
But if I want to select the last element with df.col2[-1] it results in the error message:
KeyError: -1
I know that there are workarounds to that. I could do for example df.col2[len(df)-1] or df.iloc[-1,1]. But why wouldn't be the much simpler version of indexing directly by -1 be allowed? Am I maybe missing another simple selection way for -1? Tnx
The index labels of your DataFrame are [0,1,2]. Your code df.col2[1] is an equivalent of using a loc function as df['col2'].loc[1](or df.col2.loc[1]). You can see that you index does not contain a label '-1' (which is why you get the KeyError).
For positional indexing you need to use an iloc function (which you can use on Pandas Series as well as DataFrame), so you could do df['col2'].iloc[-1] (or df.col2.iloc[-1]).
As you can see, you can use both label based ('col2') and position based (-1) indexing together, you don't need to choose one or another as df.iloc[-1,1] or df.col2[len(df)-1] (which would be equivalent to df.loc[lend(df)-1,'col2'])

Creating array from existing array based on conditions of other arrays

itmPaths is a list with 719255 (integer) values.
Pt is a 719255x1 matrix/array with float64 values.
C is a 719255x1 matrix/array with float64 values.
I would like to extract the index values where Pt > C, and then use those index values to extract the values from itmPaths that correspond with those index values, and then store those values in a new array, called exPaths. I have tried using the following code:
exPaths = itmPaths[index for index,value in enumerate(Pt-C) if value > 0]
In Matlab I can successfully do this using:
exPaths = itmPaths(Pt>C);
I would like to keep the code as efficient as possible. Thanks.
Using a list comprehension, you could do this, but just as I don't know the exact structure of what you call matrix you may adapt, but zipping both allows to keep track of the index (to extract value after) and the values (to apply the condition)
exPaths = [itmPaths[idx] for idx, pc in enumerate(zip(Pt, C)) if pc[0] > pc[1]]

Compare each column to a different value efficiently

I have a numpy array of 4000*6 (6 column). And I have a numpy column (1*6) of minimum values (made from another numpy array of 3000*6).
I want to find everything in the large array that is below those values. but each value to it's corresponding column.
I've tried the simple way, based on a one column solution I already had:
largearray=[float('nan') if x<min_values else x for x in largearray]
but sadly it didn't work :(.
I can do a for loop for each column and each value, but i was wondering if there is a faster more elegant solution.
Thanks
EDIT: I'll try to rephrase: I have 6 values, and 6 columns.
i want to find the values in each column that are lower then the corresponding one from the 6 values.
by array I mean a 2d array. sorry if it wasn't clear
sorry, i'm still thinking in Matlab a bit.
this my loop solution. It's on df, not numpy. still, is there a faster way?
a=0
for y in dfnames:
df[y]=[float('nan') if x<minvalues[a] else x for x in df[y]]
a=a+1
df is the large array or dataframe
dfnames are the column names i'm interested in.
minvalues are the minimum values for each column. I'm assuming that the order is the same. bad assumption, but works for now.
will appreciate any help making it better
I think you just need
result = largearray.copy()
result[result < min_values] = np.nan
That is, result is a copy of largearray but ay element less than the corresponding column of min_values is set to nan.
If you want to blank entire rows only when all entries in the row are less than the corresponding column of min_values, then you want:
result = largearray.copy()
result[np.all(result < min_values, axis=1)] = np.nan
I don't use numpy, so it may be not commont used solution, but such work:
largearray = numpy.array([[1,2,3], [3,4,5]])
minvalues =numpy.array([3,4,5])
largearray1=[(float('nan') if not numpy.all(numpy.less(x, min_values)) else x) for x in largearray]
result should be: [[1,2,3], 'nan']

How to return the index value of an element in a pandas dataframe

I have a dataframe of corporate actions for a specific equity. it looks something like this:
0 Declared Date Ex-Date Record Date
BAR_DATE
2018-01-17 2017-02-21 2017-08-09 2017-08-11
2018-01-16 2017-02-21 2017-05-10 2017-06-05
except that it has hundreds of rows, but that is unimportant. I created the index "BAR_DATE" from one of the columns which is where the 0 comes from above BAR_DATE.
What I want to do is to be able to reference a specific element of the dataframe and return the index value, or BAR_DATE, I think it would go something like this:
index_value = cacs.iloc[5, :].index.get_values()
except index_value becomes the column names, not the index. Now, this may stem from a poor understanding of indexing in pandas dataframes, so this may or may not be really easy to solve for someone else.
I have looked at a number of other questions including this one, but it returns column values as well.
Your code is really close, but you took it just one step further than you needed to.
# creates a slice of the dataframe where the row is at iloc 5 (row number 5) and where the slice includes all columns
slice_of_df = cacs.iloc[5, :]
# returns the index of the slice
# this will be an Index() object
index_of_slice = slice_of_df.index
From here we can use the documentation on the Index object: https://pandas.pydata.org/pandas-docs/stable/generated/pandas.Index.html
# turns the index into a list of values in the index
index_list = index_of_slice.to_list()
# gets the first index value
first_value = index_list[0]
The most important thing to remember about the Index is that it is an object of its own, and thus we need to change it to the type we expect to work with if we want something other than an index. This is where documentation can be a huge help.
EDIT: It turns out that the iloc in this case is returning a Series object which is why the solution is returning the wrong value. Knowing this, the new solution would be:
# creates a Series object from row 5 (technically the 6th row)
row_as_series = cacs.iloc[5, :]
# the name of a series relates to it's index
index_of_series = row_as_series.name
This would be the approach for single-row indexing. You would use the former approach with multi-row indexing where the return value is a DataFrame and not a Series.
Unfortunately, I don't know how to coerce the Series into a DataFrame for single-row slicingbeyond explicit conversion:
row_as_df = DataFrame(cacs.iloc[5, :])
While this will work, and the first approach will happily take this and return the index, there is likely a reason why Pandas doesn't return a DataFrame for single-row slicing so I am hesitant to offer this as a solution.

What is Pandas doing here that my indexes [0] and [1] refer to the same value?

I have a dataframe with these indices and values:
df[df.columns[0]]
1 example
2 example1
3 example2
When I access df[df.columns[0]][2], I get "example1". Makes sense. That's how indices work.
When I access df[df.columns[0]], however, I get "example", and I get example when I access df[df.columns[1]] as well. So for
df[df.columns[0]][0]
df[df.columns[0]][1]
I get "example".
Strangely, I can delete "row" 0, and the result is that 1 is deleted:
gf = df.drop(df.index[[0]])
gf
exampleDF
2 example1
3 example2
But when I delete row 1, then
2 example1
is deleted, as opposed to example.
This is a bit confusing to me; are there inconsistent standards in Pandas regarding row indices, or am I missing something / made an error?
You are probably causing pandas to switch between .iloc (index based) and .loc (label based) indexing.
All arrays in Python are 0 indexed. And I notice that indexes in your DataFrame are starting from 1. So when you run df[df.column[0]][0] pandas realizes that there is no index named 0, and falls back to .iloc which locates things by array indexing. Therefore it returns what it finds at the first location of the array, which is 'example'.
When you run df[df.column[0]][1] however, pandas realizes that there is a index label 1, and uses .loc which returns what it finds at that label, which again happens to be 'example'.
When you delete the first row, your DataFrame does not have index labels 0 and 1. So when you go to locate elements at those places in the way you are, it does not return None to you, but instead falls back on array based indexing and returns elements from the 0th and 1st places in the array.
To enforce pandas to use one of the two indexing techniques, use .iloc or .loc. .loc is label based, and will raise KeyError if you try df[df.column[0]].loc[0]. .iloc is index based and will return 'example' when you try df[df.column[0]].iloc[0].
Additional note
These commands are bad practice: df[col_label].iloc[row_index]; df[col_label].loc[row_label].
Please use df.loc[row_label, col_label]; or df.iloc[row_index, col_index]; or df.ix[row_label_or_index, col_label_or_index]
See Different Choices for Indexing for more information.

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