We Keep Coding

Why is that Pandas to numpy array conversion yields different results with the inclusion of brackets [duplicate]

I'm confused about the syntax regarding the following line of code:
x_values = dataframe[['Brains']]
The dataframe object consists of 2 columns (Brains and Bodies)
Brains Bodies
42 34
32 23
When I print x_values I get something like this:
Brains
0 42
1 32
I'm aware of the pandas documentation as far as attributes and methods of the dataframe object are concerned, but the double bracket syntax is confusing me.

Consider this:
Source DF:
In [79]: df
Out[79]:
Brains Bodies
0 42 34
1 32 23
Selecting one column - results in Pandas.Series:
In [80]: df['Brains']
Out[80]:
0 42
1 32
Name: Brains, dtype: int64
In [81]: type(df['Brains'])
Out[81]: pandas.core.series.Series
Selecting subset of DataFrame - results in DataFrame:
In [82]: df[['Brains']]
Out[82]:
Brains
0 42
1 32
In [83]: type(df[['Brains']])
Out[83]: pandas.core.frame.DataFrame
Conclusion: the second approach allows us to select multiple columns from the DataFrame. The first one just for selecting single column...
Demo:
In [84]: df = pd.DataFrame(np.random.rand(5,6), columns=list('abcdef'))
In [85]: df
Out[85]:
a b c d e f
0 0.065196 0.257422 0.273534 0.831993 0.487693 0.660252
1 0.641677 0.462979 0.207757 0.597599 0.117029 0.429324
2 0.345314 0.053551 0.634602 0.143417 0.946373 0.770590
3 0.860276 0.223166 0.001615 0.212880 0.907163 0.437295
4 0.670969 0.218909 0.382810 0.275696 0.012626 0.347549
In [86]: df[['e','a','c']]
Out[86]:
e a c
0 0.487693 0.065196 0.273534
1 0.117029 0.641677 0.207757
2 0.946373 0.345314 0.634602
3 0.907163 0.860276 0.001615
4 0.012626 0.670969 0.382810
and if we specify only one column in the list we will get a DataFrame with one column:
In [87]: df[['e']]
Out[87]:
e
0 0.487693
1 0.117029
2 0.946373
3 0.907163
4 0.012626

There is no special syntax in Python for [[ and ]]. Rather, a list is being created, and then that list is being passed as an argument to the DataFrame indexing function.
As per #MaxU's answer, if you pass a single string to a DataFrame a series that represents that one column is returned. If you pass a list of strings, then a DataFrame that contains the given columns is returned.
So, when you do the following
# Print "Brains" column as Series
print(df['Brains'])
# Return a DataFrame with only one column called "Brains"
print(df[['Brains']])
It is equivalent to the following
# Print "Brains" column as Series
column_to_get = 'Brains'
print(df[column_to_get])
# Return a DataFrame with only one column called "Brains"
subset_of_columns_to_get = ['Brains']
print(df[subset_of_columns_to_get])
In both cases, the DataFrame is being indexed with the [] operator.
Python uses the [] operator for both indexing and for constructing list literals, and ultimately I believe this is your confusion. The outer [ and ] in df[['Brains']] is performing the indexing, and the inner is creating a list.
>>> some_list = ['Brains']
>>> some_list_of_lists = [['Brains']]
>>> ['Brains'] == [['Brains']][0]
True
>>> 'Brains' == [['Brains']][0][0] == [['Brains'][0]][0]
True
What I am illustrating above is that at no point does Python ever see [[ and interpret it specially. In the last convoluted example ([['Brains'][0]][0]) there is no special ][ operator or ]][ operator... what happens is
A single-element list is created (['Brains'])
The first element of that list is indexed (['Brains'][0] => 'Brains')
That is placed into another list ([['Brains'][0]] => ['Brains'])
And then the first element of that list is indexed ([['Brains'][0]][0] => 'Brains')

Other solutions demonstrate the difference between a series and a dataframe. For the Mathematically minded, you may wish to consider the dimensions of your input and output. Here's a summary:
Object Series DataFrame
Dimensions (obj.ndim) 1 2
Syntax arg dim 0 1
Syntax df['col'] df[['col']]
Max indexing dim 1 2
Label indexing df['col'].loc[x] df.loc[x, 'col']
Label indexing (scalar) df['col'].at[x] df.at[x, 'col']
Integer indexing df['col'].iloc[x] df.iloc[x, 'col']
Integer indexing (scalar) df['col'].iat[x] dfi.at[x, 'col']
When you specify a scalar or list argument to pd.DataFrame.__getitem__, for which [] is syntactic sugar, the dimension of your argument is one less than the dimension of your result. So a scalar (0-dimensional) gives a 1-dimensional series. A list (1-dimensional) gives a 2-dimensional dataframe. This makes sense since the additional dimension is the dataframe index, i.e. rows. This is the case even if your dataframe happens to have no rows.

[ ] and [[ ]] are the concept of NumPy.
Try to understand the basics of np.array creating and use reshape and check with ndim, you'll understand.
Check my answer here.
https://stackoverflow.com/a/70194733/7660981

Python pandas, how can I pass a colon ":" to an indexer through a variable

I am working through a method that ultimately will be working with data slices from a large multi-index pandas dataframe. I can generate masks to use for each indexer (essentially lists of values to define the slice):
df.loc[idx[a_mask,b_mask],idx[c_mask,d_mask]]
This would be fine but in some scenarios I'd really like to select everything along some of those axes, something equivalent to:
df.loc[idx[a_mask,b_mask],idx[:,d_mask]]
Is there a way for me to pass that colon ":" that replaces the c_mask in the second example in as a variable? Ideally I'd just set the c_mask to a value like ":", but of course that doesn't work (and shouldn't because what if we had a column named that...). But is there any way to pass in a value by variable that communicates "whole axis" along one of those indexers?
I do realize I could generate a mask that would select everything by gathering together all the values along the appropriate axis, but this is nontrivial and adds a lot of code. Likewise I could break the dataframe access into 5 scenarios (one each for having a : in it and one with four masks) but that doesn't seem to honor the DRY principle and is still brittle because it can't handle multiple direction whole slice selection.
So, anything I can pass in through a variable that will select an entire direction in an indexer like a : would? Or is there a more elegant way to optionally select an entire direction?

idx[slice(None)] is equivalent to the idx[:]
So these are all equivalent.
In [11]: df = DataFrame({'A' : np.random.randn(9)},index=pd.MultiIndex.from_product([range(3),list('abc')],names=['first','second']))
In [12]: df
Out[12]:
A
first second
0 a -0.668344
b -1.679159
c 0.061876
1 a -0.237272
b 0.136495
c -1.296027
2 a 0.554533
b 0.433941
c -0.014107
In [13]: idx = pd.IndexSlice
In [14]: df.loc[idx[:,'b'],]
Out[14]:
A
first second
0 b -1.679159
1 b 0.136495
2 b 0.433941
In [15]: df.loc[idx[slice(None),'b'],]
Out[15]:
A
first second
0 b -1.679159
1 b 0.136495
2 b 0.433941
In [16]: df.loc[(slice(None),'b'),]
Out[16]:
A
first second
0 b -1.679159
1 b 0.136495
2 b 0.433941

Python Pandas: Resolving "List Object has no Attribute 'Loc'"

I import a CSV as a DataFrame using:
import numpy as np
import pandas as pd
df = pd.read_csv("test.csv")
Then I'm trying to do a simple replace based on IDs:
df.loc[df.ID == 103, ['fname', 'lname']] = 'Michael', 'Johnson'
I get the following error:
AttributeError: 'list' object has no attribute 'loc'
Note, when I do print pd.version() I get 0.12.0, so it's not a problem (at least as far as I understand) with having pre-11 version. Any ideas?

To pickup from the comment: "I was doing this:"
df = [df.hc== 2]
What you create there is a "mask": an array with booleans that says which part of the index fulfilled your condition.
To filter your dataframe on your condition you want to do this:
df = df[df.hc == 2]
A bit more explicit is this:
mask = df.hc == 2
df = df[mask]
If you want to keep the entire dataframe and only want to replace specific values, there are methods such replace: Python pandas equivalent for replace. Also another (performance wise great) method would be creating a separate DataFrame with the from/to values as column and using pd.merge to combine it into the existing DataFrame. And using your index to set values is also possible:
df[mask]['fname'] = 'Johnson'
But for a larger set of replaces you would want to use one of the two other methods or use "apply" with a lambda function (for value transformations). Last but not least: you can use .fillna('bla') to rapidly fill up NA values.

The traceback indicates to you that df is a list and not a DataFrame as expected in your line of code.
It means that between df = pd.read_csv("test.csv") and df.loc[df.ID == 103, ['fname', 'lname']] = 'Michael', 'Johnson' you have other lines of codes that assigns a list object to df. Review that piece of code to find your bug

#Boud answer is correct. Loc assignment works fine if the right-hand-side list matches the number of replacing elements
In [56]: df = DataFrame(dict(A =[1,2,3], B = [4,5,6], C = [7,8,9]))
In [57]: df
Out[57]:
A B C
0 1 4 7
1 2 5 8
2 3 6 9
In [58]: df.loc[1,['A','B']] = -1,-2
In [59]: df
Out[59]:
A B C
0 1 4 7
1 -1 -2 8
2 3 6 9

Adding calculated column(s) to a dataframe in pandas

I have an OHLC price data set, that I have parsed from CSV into a Pandas dataframe and resampled to 15 min bars:
<class 'pandas.core.frame.DataFrame'>
DatetimeIndex: 500047 entries, 1998-05-04 04:45:00 to 2012-08-07 00:15:00
Freq: 15T
Data columns:
Close 363152 non-null values
High 363152 non-null values
Low 363152 non-null values
Open 363152 non-null values
dtypes: float64(4)
I would like to add various calculated columns, starting with simple ones such as period Range (H-L) and then booleans to indicate the occurrence of price patterns that I will define - e.g. a hammer candle pattern, for which a sample definition:
def closed_in_top_half_of_range(h,l,c):
return c > l + (h-l)/2
def lower_wick(o,l,c):
return min(o,c)-l
def real_body(o,c):
return abs(c-o)
def lower_wick_at_least_twice_real_body(o,l,c):
return lower_wick(o,l,c) >= 2 * real_body(o,c)
def is_hammer(row):
return lower_wick_at_least_twice_real_body(row["Open"],row["Low"],row["Close"]) \
and closed_in_top_half_of_range(row["High"],row["Low"],row["Close"])
Basic problem: how do I map the function to the column, specifically where I would like to reference more than one other column or the whole row or whatever?
This post deals with adding two calculated columns off of a single source column, which is close, but not quite it.
And slightly more advanced: for price patterns that are determined with reference to more than a single bar (T), how can I reference different rows (e.g. T-1, T-2 etc.) from within the function definition?

The exact code will vary for each of the columns you want to do, but it's likely you'll want to use the map and apply functions. In some cases you can just compute using the existing columns directly, since the columns are Pandas Series objects, which also work as Numpy arrays, which automatically work element-wise for usual mathematical operations.
>>> d
A B C
0 11 13 5
1 6 7 4
2 8 3 6
3 4 8 7
4 0 1 7
>>> (d.A + d.B) / d.C
0 4.800000
1 3.250000
2 1.833333
3 1.714286
4 0.142857
>>> d.A > d.C
0 True
1 True
2 True
3 False
4 False
If you need to use operations like max and min within a row, you can use apply with axis=1 to apply any function you like to each row. Here's an example that computes min(A, B)-C, which seems to be like your "lower wick":
>>> d.apply(lambda row: min([row['A'], row['B']])-row['C'], axis=1)
0 6
1 2
2 -3
3 -3
4 -7
Hopefully that gives you some idea of how to proceed.
Edit: to compare rows against neighboring rows, the simplest approach is to slice the columns you want to compare, leaving off the beginning/end, and then compare the resulting slices. For instance, this will tell you for which rows the element in column A is less than the next row's element in column C:
d['A'][:-1] < d['C'][1:]
and this does it the other way, telling you which rows have A less than the preceding row's C:
d['A'][1:] < d['C'][:-1]
Doing ['A"][:-1] slices off the last element of column A, and doing ['C'][1:] slices off the first element of column C, so when you line these two up and compare them, you're comparing each element in A with the C from the following row.

You could have is_hammer in terms of row["Open"] etc. as follows
def is_hammer(rOpen,rLow,rClose,rHigh):
return lower_wick_at_least_twice_real_body(rOpen,rLow,rClose) \
and closed_in_top_half_of_range(rHigh,rLow,rClose)
Then you can use map:
df["isHammer"] = map(is_hammer, df["Open"], df["Low"], df["Close"], df["High"])

For the second part of your question, you can also use shift, for example:
df['t-1'] = df['t'].shift(1)
t-1 would then contain the values from t one row above.
http://pandas.pydata.org/pandas-docs/stable/generated/pandas.DataFrame.shift.html

The first four functions you list will work on vectors as well, with the exception that lower_wick needs to be adapted. Something like this,
def lower_wick_vec(o, l, c):
min_oc = numpy.where(o > c, c, o)
return min_oc - l
where o, l and c are vectors.
You could do it this way instead which just takes the df as input and avoid using numpy, although it will be much slower:
def lower_wick_df(df):
min_oc = df[['Open', 'Close']].min(axis=1)
return min_oc - l
The other three will work on columns or vectors just as they are. Then you can finish off with
def is_hammer(df):
lw = lower_wick_at_least_twice_real_body(df["Open"], df["Low"], df["Close"])
cl = closed_in_top_half_of_range(df["High"], df["Low"], df["Close"])
return cl & lw
Bit operators can perform set logic on boolean vectors, & for and, | for or etc. This is enough to completely vectorize the sample calculations you gave and should be relatively fast. You could probably speed up even more by temporarily working with the numpy arrays underlying the data while performing these calculations.
For the second part, I would recommend introducing a column indicating the pattern for each row and writing a family of functions which deal with each pattern. Then groupby the pattern and apply the appropriate function to each group.

Find row where values for column is maximal in a pandas DataFrame

How can I find the row for which the value of a specific column is maximal?
df.max() will give me the maximal value for each column, I don't know how to get the corresponding row.

Use the pandas idxmax function. It's straightforward:
>>> import pandas
>>> import numpy as np
>>> df = pandas.DataFrame(np.random.randn(5,3),columns=['A','B','C'])
>>> df
A B C
0 1.232853 -1.979459 -0.573626
1 0.140767 0.394940 1.068890
2 0.742023 1.343977 -0.579745
3 2.125299 -0.649328 -0.211692
4 -0.187253 1.908618 -1.862934
>>> df['A'].idxmax()
3
>>> df['B'].idxmax()
4
>>> df['C'].idxmax()
1
Alternatively you could also use numpy.argmax, such as numpy.argmax(df['A']) -- it provides the same thing, and appears at least as fast as idxmax in cursory observations.
idxmax() returns indices labels, not integers.
Example': if you have string values as your index labels, like rows 'a' through 'e', you might want to know that the max occurs in row 4 (not row 'd').
if you want the integer position of that label within the Index you have to get it manually (which can be tricky now that duplicate row labels are allowed).
HISTORICAL NOTES:
idxmax() used to be called argmax() prior to 0.11
argmax was deprecated prior to 1.0.0 and removed entirely in 1.0.0
back as of Pandas 0.16, argmax used to exist and perform the same function (though appeared to run more slowly than idxmax).
argmax function returned the integer position within the index of the row location of the maximum element.
pandas moved to using row labels instead of integer indices. Positional integer indices used to be very common, more common than labels, especially in applications where duplicate row labels are common.
For example, consider this toy DataFrame with a duplicate row label:
In [19]: dfrm
Out[19]:
A B C
a 0.143693 0.653810 0.586007
b 0.623582 0.312903 0.919076
c 0.165438 0.889809 0.000967
d 0.308245 0.787776 0.571195
e 0.870068 0.935626 0.606911
f 0.037602 0.855193 0.728495
g 0.605366 0.338105 0.696460
h 0.000000 0.090814 0.963927
i 0.688343 0.188468 0.352213
i 0.879000 0.105039 0.900260
In [20]: dfrm['A'].idxmax()
Out[20]: 'i'
In [21]: dfrm.iloc[dfrm['A'].idxmax()] # .ix instead of .iloc in older versions of pandas
Out[21]:
A B C
i 0.688343 0.188468 0.352213
i 0.879000 0.105039 0.900260
So here a naive use of idxmax is not sufficient, whereas the old form of argmax would correctly provide the positional location of the max row (in this case, position 9).
This is exactly one of those nasty kinds of bug-prone behaviors in dynamically typed languages that makes this sort of thing so unfortunate, and worth beating a dead horse over. If you are writing systems code and your system suddenly gets used on some data sets that are not cleaned properly before being joined, it's very easy to end up with duplicate row labels, especially string labels like a CUSIP or SEDOL identifier for financial assets. You can't easily use the type system to help you out, and you may not be able to enforce uniqueness on the index without running into unexpectedly missing data.
So you're left with hoping that your unit tests covered everything (they didn't, or more likely no one wrote any tests) -- otherwise (most likely) you're just left waiting to see if you happen to smack into this error at runtime, in which case you probably have to go drop many hours worth of work from the database you were outputting results to, bang your head against the wall in IPython trying to manually reproduce the problem, finally figuring out that it's because idxmax can only report the label of the max row, and then being disappointed that no standard function automatically gets the positions of the max row for you, writing a buggy implementation yourself, editing the code, and praying you don't run into the problem again.

You might also try idxmax:
In [5]: df = pandas.DataFrame(np.random.randn(10,3),columns=['A','B','C'])
In [6]: df
Out[6]:
A B C
0 2.001289 0.482561 1.579985
1 -0.991646 -0.387835 1.320236
2 0.143826 -1.096889 1.486508
3 -0.193056 -0.499020 1.536540
4 -2.083647 -3.074591 0.175772
5 -0.186138 -1.949731 0.287432
6 -0.480790 -1.771560 -0.930234
7 0.227383 -0.278253 2.102004
8 -0.002592 1.434192 -1.624915
9 0.404911 -2.167599 -0.452900
In [7]: df.idxmax()
Out[7]:
A 0
B 8
C 7
e.g.
In [8]: df.loc[df['A'].idxmax()]
Out[8]:
A 2.001289
B 0.482561
C 1.579985

Both above answers would only return one index if there are multiple rows that take the maximum value. If you want all the rows, there does not seem to have a function.
But it is not hard to do. Below is an example for Series; the same can be done for DataFrame:
In [1]: from pandas import Series, DataFrame
In [2]: s=Series([2,4,4,3],index=['a','b','c','d'])
In [3]: s.idxmax()
Out[3]: 'b'
In [4]: s[s==s.max()]
Out[4]:
b 4
c 4
dtype: int64

df.iloc[df['columnX'].argmax()]
argmax() would provide the index corresponding to the max value for the columnX. iloc can be used to get the row of the DataFrame df for this index.

A more compact and readable solution using query() is like this:
import pandas as pd
df = pandas.DataFrame(np.random.randn(5,3),columns=['A','B','C'])
print(df)
# find row with maximum A
df.query('A == A.max()')
It also returns a DataFrame instead of Series, which would be handy for some use cases.

Very simple: we have df as below and we want to print a row with max value in C:
A B C
x 1 4
y 2 10
z 5 9
In:
df.loc[df['C'] == df['C'].max()] # condition check
Out:
A B C
y 2 10

If you want the entire row instead of just the id, you can use df.nlargest and pass in how many 'top' rows you want and you can also pass in for which column/columns you want it for.
df.nlargest(2,['A'])
will give you the rows corresponding to the top 2 values of A.
use df.nsmallest for min values.

The direct ".argmax()" solution does not work for me.
The previous example provided by #ely
>>> import pandas
>>> import numpy as np
>>> df = pandas.DataFrame(np.random.randn(5,3),columns=['A','B','C'])
>>> df
A B C
0 1.232853 -1.979459 -0.573626
1 0.140767 0.394940 1.068890
2 0.742023 1.343977 -0.579745
3 2.125299 -0.649328 -0.211692
4 -0.187253 1.908618 -1.862934
>>> df['A'].argmax()
3
>>> df['B'].argmax()
4
>>> df['C'].argmax()
1
returns the following message :
FutureWarning: 'argmax' is deprecated, use 'idxmax' instead. The behavior of 'argmax'
will be corrected to return the positional maximum in the future.
Use 'series.values.argmax' to get the position of the maximum now.
So that my solution is :
df['A'].values.argmax()

mx.iloc[0].idxmax()
This one line of code will give you how to find the maximum value from a row in dataframe, here mx is the dataframe and iloc[0] indicates the 0th index.

Considering this dataframe
[In]: df = pd.DataFrame(np.random.randn(4,3),columns=['A','B','C'])
[Out]:
A B C
0 -0.253233 0.226313 1.223688
1 0.472606 1.017674 1.520032
2 1.454875 1.066637 0.381890
3 -0.054181 0.234305 -0.557915
Assuming one want to know the rows where column "C" is max, the following will do the work
[In]: df[df['C']==df['C'].max()])
[Out]:
A B C
1 0.472606 1.017674 1.520032

The idmax of the DataFrame returns the label index of the row with the maximum value and the behavior of argmax depends on version of pandas (right now it returns a warning). If you want to use the positional index, you can do the following:
max_row = df['A'].values.argmax()
or
import numpy as np
max_row = np.argmax(df['A'].values)
Note that if you use np.argmax(df['A']) behaves the same as df['A'].argmax().

Use:
data.iloc[data['A'].idxmax()]
data['A'].idxmax() -finds max value location in terms of row
data.iloc() - returns the row

If there are ties in the maximum values, then idxmax returns the index of only the first max value. For example, in the following DataFrame:
A B C
0 1 0 1
1 0 0 1
2 0 0 0
3 0 1 1
4 1 0 0
idxmax returns
A 0
B 3
C 0
dtype: int64
Now, if we want all indices corresponding to max values, then we could use max + eq to create a boolean DataFrame, then use it on df.index to filter out indexes:
out = df.eq(df.max()).apply(lambda x: df.index[x].tolist())
Output:
A [0, 4]
B [3]
C [0, 1, 3]
dtype: object

what worked for me is:
df[df['colX'] == df['colX'].max()
You then get the row in your df with the maximum value of colX.
Then if you just want the index you can add .index at the end of the query.