We Keep Coding

Normalization of a matrix

I have a 150x4 matrix X which I created from a pandas dataframe using the following code:
X = df_new.as_matrix()
I have to normalize it using this function:
I know that Uj is the mean val of j, and that σ j is the standard deviation of j, but I don't understand what j is. I'm having a little trouble understanding what the bar on X is, and I'm confused by the commas in the equation (I don't know if they have any significance or not).
Can anyone help me understand what this equation means so I can then write the normalization using sklearn?

You don't actually need to write code for the normalization yourself - it comes ready with sklearn.preprocessing.scale.
Here is an example from the docs:
>>> from sklearn import preprocessing
>>> import numpy as np
>>> X_train = np.array([[ 1., -1., 2.],
... [ 2., 0., 0.],
... [ 0., 1., -1.]])
>>> X_scaled = preprocessing.scale(X_train)
>>> X_scaled
array([[ 0. ..., -1.22..., 1.33...],
[ 1.22..., 0. ..., -0.26...],
[-1.22..., 1.22..., -1.06...]])
When used with the default setting axis=0, the mormalization happens column-wise (i.e. for each column j, as in your equestion). As a result, it is easy to confirm that scaled data has zero mean and unit variance:
>>> X_scaled.mean(axis=0)
array([ 0., 0., 0.])
>>> X_scaled.std(axis=0)
array([ 1., 1., 1.])

The indexes for matrix X are row (i) and column (j). Hence, X,j means column j of matrix X. I.e. normalize each column of matrix X to z-scores.
You can do that using pandas:
df_new_zscores = (df_new - df_new.mean()) / df_new.std()

I do not know pandas but I think that the equation means that the normalized matrix is given by
You subtract the empirical mean and devide by the empirical standard deviation per column.
You sometimes use this for Principal Component Analysis.

Eigenvectors computed with numpy's eigh and svd do not match

Consider singular value decomposition M=USV*. Then the eigenvalue decomposition of M* M gives M* M= V (S* S) V*=VS* U* USV*. I wish to verify this equality with numpy by showing that the eigenvectors returned by eigh function are the same as those returned by svd function:
import numpy as np
np.random.seed(42)
# create mean centered data
A=np.random.randn(50,20)
M= A-np.array(A.mean(0),ndmin=2)
# svd
U1,S1,V1=np.linalg.svd(M)
S1=np.square(S1)
V1=V1.T
# eig
S2,V2=np.linalg.eigh(np.dot(M.T,M))
indx=np.argsort(S2)[::-1]
S2=S2[indx]
V2=V2[:,indx]
# both Vs are in orthonormal form
assert np.all(np.isclose(np.linalg.norm(V1,axis=1), np.ones(V1.shape[0])))
assert np.all(np.isclose(np.linalg.norm(V1,axis=0), np.ones(V1.shape[1])))
assert np.all(np.isclose(np.linalg.norm(V2,axis=1), np.ones(V2.shape[0])))
assert np.all(np.isclose(np.linalg.norm(V2,axis=0), np.ones(V2.shape[1])))
assert np.all(np.isclose(S1,S2))
assert np.all(np.isclose(V1,V2))
The last assertion fails. Why?

Just play with small numbers to debug your problem.
Start with A=np.random.randn(3,2) instead of your much larger matrix with size (50,20)
In my random case, I find that
v1 = array([[-0.33872745, 0.94088454],
[-0.94088454, -0.33872745]])
and for v2:
v2 = array([[ 0.33872745, -0.94088454],
[ 0.94088454, 0.33872745]])
they only differ for a sign, and obviously, even if normalized to have unit module, the vector can differ for a sign.
Now if you try the trick
assert np.all(np.isclose(V1,-1*V2))
for your original big matrix, it fails... again, this is OK. What happens is that some vectors have been multiplied by -1, some others haven't.
A correct way to check for equality between the vectors is:
assert allclose(abs((V1*V2).sum(0)),1.)
and indeed, to get a feeling of how this works you can print this quantity:
(V1*V2).sum(0)
that indeed is either +1 or -1 depending on the vector:
array([ 1., -1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1., 1.,
1., -1., 1., 1., 1., -1., -1.])
EDIT: This will happen in most cases, especially if starting from a random matrix. Notice however that this test will likely fail if one or more eigenvalues has an eigenspace of dimension larger than 1, as pointed out by #Sven Marnach in his comment below:
There might be other differences than just vectors multiplied by -1.
If any of the eigenvalues has a multi-dimensional eigenspace, you
might get an arbitrary orthonormal basis of that eigenspace, and to
such bases might be rotated against each other by an arbitraty
unitarian matrix

What does axis in pandas mean?

Here is my code to generate a dataframe:
import pandas as pd
import numpy as np
dff = pd.DataFrame(np.random.randn(1,2),columns=list('AB'))
then I got the dataframe:
+------------+---------+--------+
| | A | B |
+------------+---------+---------
| 0 | 0.626386| 1.52325|
+------------+---------+--------+
When I type the commmand :
dff.mean(axis=1)
I got :
0 1.074821
dtype: float64
According to the reference of pandas, axis=1 stands for columns and I expect the result of the command to be
A 0.626386
B 1.523255
dtype: float64
So here is my question: what does axis in pandas mean?

It specifies the axis along which the means are computed. By default axis=0. This is consistent with the numpy.mean usage when axis is specified explicitly (in numpy.mean, axis==None by default, which computes the mean value over the flattened array) , in which axis=0 along the rows (namely, index in pandas), and axis=1 along the columns. For added clarity, one may choose to specify axis='index' (instead of axis=0) or axis='columns' (instead of axis=1).
+------------+---------+--------+
| | A | B |
+------------+---------+---------
| 0 | 0.626386| 1.52325|----axis=1----->
+------------+---------+--------+
| |
| axis=0 |
↓ ↓

These answers do help explain this, but it still isn't perfectly intuitive for a non-programmer (i.e. someone like me who is learning Python for the first time in context of data science coursework). I still find using the terms "along" or "for each" wrt to rows and columns to be confusing.
What makes more sense to me is to say it this way:
Axis 0 will act on all the ROWS in each COLUMN
Axis 1 will act on all the COLUMNS in each ROW
So a mean on axis 0 will be the mean of all the rows in each column, and a mean on axis 1 will be a mean of all the columns in each row.
Ultimately this is saying the same thing as #zhangxaochen and #Michael, but in a way that is easier for me to internalize.

Let's visualize (you gonna remember always),
In Pandas:
axis=0 means along "indexes". It's a row-wise operation.
Suppose, to perform concat() operation on dataframe1 & dataframe2,
we will take dataframe1 & take out 1st row from dataframe1 and place into the new DF, then we take out another row from dataframe1 and put into new DF, we repeat this process until we reach to the bottom of dataframe1. Then, we do the same process for dataframe2.
Basically, stacking dataframe2 on top of dataframe1 or vice a versa.
E.g making a pile of books on a table or floor
axis=1 means along "columns". It's a column-wise operation.
Suppose, to perform concat() operation on dataframe1 & dataframe2,
we will take out the 1st complete column(a.k.a 1st series) of dataframe1 and place into new DF, then we take out the second column of dataframe1 and keep adjacent to it (sideways), we have to repeat this operation until all columns are finished. Then, we repeat the same process on dataframe2.
Basically,
stacking dataframe2 sideways.
E.g arranging books on a bookshelf.
More to it, since arrays are better representations to represent a nested n-dimensional structure compared to matrices! so below can help you more to visualize how axis plays an important role when you generalize to more than one dimension. Also, you can actually print/write/draw/visualize any n-dim array but, writing or visualizing the same in a matrix representation(3-dim) is impossible on a paper more than 3-dimensions.

axis refers to the dimension of the array, in the case of pd.DataFrames axis=0 is the dimension that points downwards and axis=1 the one that points to the right.
Example: Think of an ndarray with shape (3,5,7).
a = np.ones((3,5,7))
a is a 3 dimensional ndarray, i.e. it has 3 axes ("axes" is plural of "axis"). The configuration of a will look like 3 slices of bread where each slice is of dimension 5-by-7. a[0,:,:] will refer to the 0-th slice, a[1,:,:] will refer to the 1-st slice etc.
a.sum(axis=0) will apply sum() along the 0-th axis of a. You will add all the slices and end up with one slice of shape (5,7).
a.sum(axis=0) is equivalent to
b = np.zeros((5,7))
for i in range(5):
for j in range(7):
b[i,j] += a[:,i,j].sum()
b and a.sum(axis=0) will both look like this
array([[ 3., 3., 3., 3., 3., 3., 3.],
[ 3., 3., 3., 3., 3., 3., 3.],
[ 3., 3., 3., 3., 3., 3., 3.],
[ 3., 3., 3., 3., 3., 3., 3.],
[ 3., 3., 3., 3., 3., 3., 3.]])
In a pd.DataFrame, axes work the same way as in numpy.arrays: axis=0 will apply sum() or any other reduction function for each column.
N.B. In #zhangxaochen's answer, I find the phrases "along the rows" and "along the columns" slightly confusing. axis=0 should refer to "along each column", and axis=1 "along each row".

The easiest way for me to understand is to talk about whether you are calculating a statistic for each column (axis = 0) or each row (axis = 1). If you calculate a statistic, say a mean, with axis = 0 you will get that statistic for each column. So if each observation is a row and each variable is in a column, you would get the mean of each variable. If you set axis = 1 then you will calculate your statistic for each row. In our example, you would get the mean for each observation across all of your variables (perhaps you want the average of related measures).
axis = 0: by column = column-wise = along the rows
axis = 1: by row = row-wise = along the columns

Let's look at the table from Wiki. This is an IMF estimate of GDP from 2010 to 2019 for top ten countries.
1. Axis 1 will act for each row on all the columns If you want to calculate the average (mean) GDP for EACH countries over the decade (2010-2019), you need to do, df.mean(axis=1). For example, if you want to calculate mean GDP of United States from 2010 to 2019, df.loc['United States','2010':'2019'].mean(axis=1)
2. Axis 0 will act for each column on all the rows If I want to calculate the average (mean) GDP for EACH year for all countries, you need to do, df.mean(axis=0). For example, if you want to calculate mean GDP of the year 2015 for United States, China, Japan, Germany and India, df.loc['United States':'India','2015'].mean(axis=0)
Note: The above code will work only after setting "Country(or dependent territory)" column as the Index, using set_index method.

Axis in view of programming is the position in the shape tuple. Here is an example:
import numpy as np
a=np.arange(120).reshape(2,3,4,5)
a.shape
Out[3]: (2, 3, 4, 5)
np.sum(a,axis=0).shape
Out[4]: (3, 4, 5)
np.sum(a,axis=1).shape
Out[5]: (2, 4, 5)
np.sum(a,axis=2).shape
Out[6]: (2, 3, 5)
np.sum(a,axis=3).shape
Out[7]: (2, 3, 4)
Mean on the axis will cause that dimension to be removed.
Referring to the original question, the dff shape is (1,2). Using axis=1 will change the shape to (1,).

The designer of pandas, Wes McKinney, used to work intensively on finance data. Think of columns as stock names and index as daily prices. You can then guess what the default behavior is (i.e., axis=0) with respect to this finance data. axis=1 can be simply thought as 'the other direction'.
For example, the statistics functions, such as mean(), sum(), describe(), count() all default to column-wise because it makes more sense to do them for each stock. sort_index(by=) also defaults to column. fillna(method='ffill') will fill along column because it is the same stock. dropna() defaults to row because you probably just want to discard the price on that day instead of throw away all prices of that stock.
Similarly, the square brackets indexing refers to the columns since it's more common to pick a stock instead of picking a day.

The problem with using axis= properly is for its use for 2 main different cases:
For computing an accumulated value, or rearranging (e. g. sorting) data.
For manipulating ("playing" with) entities (e. g. dataframes).
The main idea behind this answer is that for avoiding the confusion, we select either a number, or a name for specifying the particular axis, whichever is more clear, intuitive, and descriptive.
Pandas is based on NumPy, which is based on mathematics, particularly on n-dimensional matrices. Here is an image for common use of axes' names in math in the 3-dimensional space:
This picture is for memorizing the axes' ordinal numbers only:
0 for x-axis,
1 for y-axis, and
2 for z-axis.
The z-axis is only for panels; for dataframes we will restrict our interest to the green-colored, 2-dimensional basic plane with x-axis (0, vertical), and y-axis (1, horizontal).
It's all for numbers as potential values of axis= parameter.
The names of axes are 'index' (you may use the alias 'rows') and 'columns', and for this explanation it is NOT important the relation between these names and ordinal numbers (of axes), as everybody knows what the words "rows" and "columns" mean (and everybody here — I suppose — knows what the word "index" in pandas means).
And now, my recommendation:
If you want to compute an accumulated value, you may compute it from values located along axis 0 (or along axis 1) — use axis=0 (or axis=1).
Similarly, if you want to rearrange values, use the axis number of the axis, along which are located data for rearranging (e.g. for sorting).
If you want to manipulate (e.g. concatenate) entities (e.g. dataframes) — use axis='index' (synonym: axis='rows') or axis='columns' to specify the resulting change — index (rows) or columns, respectively.
(For concatenating, you will obtain either a longer index (= more rows), or more columns, respectively.)

I used to get confused with this as well, but this is how I remember it.
It specifies the dimension of the dataframe that would change or on which the operation would be performed.
Let us understand this with an example.
We have a dataframe df and it has shape as (5, 10), meaning it has 5 rows and 10 columns.
Now when we do df.mean(axis=1) it means that dimension 1 would be changed, this implies that it would have the same number of rows but a different number of columns. Hence the result that would get would be of the shape (5, 1).
Similarly, if we do df.mean(axis=0) it means that dimension 0 would be changed, meaning the number of rows would be changed but the number of columns would remain the same, hence the result would be of shape (1, 10).
Try to relate this with the examples provided in the question.

one of easy ways to remember axis 1 (columns), vs axis 0 (rows) is the output you expect.
if you expect an output for each row you use axis='columns',
on the other hand if you want an output for each column you use axis='rows'.

This is based on #Safak's answer.
The best way to understand the axes in pandas/numpy is to create a 3d array and check the result of the sum function along the 3 different axes.
a = np.ones((3,5,7))
a will be:
array([[[1., 1., 1., 1., 1., 1., 1.],
[1., 1., 1., 1., 1., 1., 1.],
[1., 1., 1., 1., 1., 1., 1.],
[1., 1., 1., 1., 1., 1., 1.],
[1., 1., 1., 1., 1., 1., 1.]],
[[1., 1., 1., 1., 1., 1., 1.],
[1., 1., 1., 1., 1., 1., 1.],
[1., 1., 1., 1., 1., 1., 1.],
[1., 1., 1., 1., 1., 1., 1.],
[1., 1., 1., 1., 1., 1., 1.]],
[[1., 1., 1., 1., 1., 1., 1.],
[1., 1., 1., 1., 1., 1., 1.],
[1., 1., 1., 1., 1., 1., 1.],
[1., 1., 1., 1., 1., 1., 1.],
[1., 1., 1., 1., 1., 1., 1.]]])
Now check out the sum of elements of the array along each of the axes:
x0 = np.sum(a,axis=0)
x1 = np.sum(a,axis=1)
x2 = np.sum(a,axis=2)
will give you the following results:
x0 :
array([[3., 3., 3., 3., 3., 3., 3.],
[3., 3., 3., 3., 3., 3., 3.],
[3., 3., 3., 3., 3., 3., 3.],
[3., 3., 3., 3., 3., 3., 3.],
[3., 3., 3., 3., 3., 3., 3.]])
x1 :
array([[5., 5., 5., 5., 5., 5., 5.],
[5., 5., 5., 5., 5., 5., 5.],
[5., 5., 5., 5., 5., 5., 5.]])
x2 :
array([[7., 7., 7., 7., 7.],
[7., 7., 7., 7., 7.],
[7., 7., 7., 7., 7.]])

I believe, the correct answer should be "it is complicated"
[1] The term "axis" iteself conjures different mental image in different people
lets say the y-axis, it should conjure up an image of something vertical. However, now think of a vertical line x=0. it is vertical line too, yet it is addressed by a value 0 on the x-axis.
Similarly, when we say axis='index' (meaning axis=0), are we saying the "vertical" direction that indexes reside on? or that one series of data addressed by an index value? Panda tends to mean the first meaning, the vertical direction.
[2] Pandas itself is not 100% consistent either, observe the following cases, they ALMOST have the same common theme:
# [1] piling dfs
pd.concat([df0, df1], axis='index')
# adding dfs on top of each other vertically like pilling up a column,
# but, we will use the word 'index'
# [2] for every column in df: operate on it
df.apply(foo, axis='index')
df.mean('A', axis='index')
a_boolean_df.all(axis='index')
# apply an operation to a vertical slice of data, ie. a column,
# then apply the same operation to the next column on the right
# then to the right again... until the last column
# but, we will use the word 'index'
# [3] delete a column or row of data
df.drop(axis='index', ...)
df.dropna(axis='index', ...)
# this time, we are droping an index/row, a horizontal slice of data.
# so OBVIOUSLY we will use the word 'index'
# [4] drop duplicate
df.drop_duplicates(subset=['mycolumn0', 'mycolumn1']...)
# thank God we don't need to deal with the "axis" bs in this

There're two most common usage of axis on Pandas:
used as indexing, like df.iloc[0, 1]
used as argument inside a function, like df.mean(axis=1)
While using as indexing, we can interpret that axis=0 stands for rows and axis=1 stands for columns, which is df.iloc[rows, columns]. So, df.iloc[0, 1] means selecting the data from row 0 and column 1, in this case, it returns 1.52325.
While using as argument, axis=0 means selecting object across rows vertically, and axis=1 means selecting object across columns horizontally.
So, df.mean(axis=1) stands for calculating mean across columns horizontally, and it returns:
0 1.074821
dtype: float64
The general purpose of axis is used for selecting specific data to operate on. And the key for understanding axis, is to separate the process of "selection" and "operation".
Let's explain it with 1 extra cases: df.drop('A', axis=1)
The operation is df.drop(), it requires the name of the intended
column, which is 'A' in this case. It's not the same as df.mean()
that operating on data content.
The selection is the name of column, not the data content of column. Since all column names are arranged across columns horizontally, so we use axis=1 to select the name object.
In short, we better separate the "selection" and "operation" to have a clear understanding on:
what object to select
how is it arranged

axis=1 ,It will give the sum row wise,keepdims=True will maintain the 2D dimension.
Hope it helps you.

Actually we don't need to hard remember what axis=0, axis=1 represents for.
As sometimes, axis can be a tuple: e.g.axis=(0,1) How do we understand such multiple dim axis?
I found if we understand how python slice [:] works, it would be easier.
Suppose we have an 1d array:
a = [ 0, 1, 0 ]
a[:] # select all the elements in array a
Suppose we have a 2d array:
M = [[0, 0, 1],
[1, 0, 0],
[0, 2, 1],
[2, 0, 2],
[3, 1, 0]]
M[1,:] # M[0]=1, M[1]=* --> [1, 0, 0]
M[:,2] # M[0]=*, M[1]=2 --> [1, 0, 1, 2, 0]
M[:,:] # M[0]=*, M[1]=* --> all the elements in M are selected
So when computes:
np.sum(M, axis=0) # [sum(M[:,0]), sum(M[:,1]), sum(M[:,2])]
np.sum(M, axis=1) # [sum(M[0,:]), sum(M[1,:]), sum(M[2,:]), sum(M[3,:]), sum(M[4,:])]
np.sum(M, axis=-1) # -1 means last dim, it's the same with np.sum(M, axis=1)
np.sum(M, axis=(0,1)) # sum(M[:,:])
The rule is simple, replace the dims specified in axis as : when computes.

axis = 0 means up to down
axis = 1 means left to right
sums[key] = lang_sets[key].iloc[:,1:].sum(axis=0)
Given example is taking sum of all the data in column == key.

My thinking : Axis = n, where n = 0, 1, etc. means that the matrix is collapsed (folded) along that axis. So in a 2D matrix, when you collapse along 0 (rows), you are really operating on one column at a time. Similarly for higher order matrices.
This is not the same as the normal reference to a dimension in a matrix, where 0 -> row and 1 -> column. Similarly for other dimensions in an N dimension array.

I'm a newbie to pandas. But this is how I understand axis in pandas:
Axis Constant Varying Direction
0 Column Row Downwards |
1 Row Column Towards Right -->
So to compute mean of a column, that particular column should be constant but the rows under that can change (varying) so it is axis=0.
Similarly, to compute mean of a row, that particular row is constant but it can traverse through different columns (varying), axis=1.

I understand this way :
Say if your operation requires traversing from left to right/right to left in a dataframe, you are apparently merging columns ie. you are operating on various columns.
This is axis =1
Example
df = pd.DataFrame(np.arange(12).reshape(3,4),columns=['A', 'B', 'C', 'D'])
print(df)
A B C D
0 0 1 2 3
1 4 5 6 7
2 8 9 10 11
df.mean(axis=1)
0 1.5
1 5.5
2 9.5
dtype: float64
df.drop(['A','B'],axis=1,inplace=True)
C D
0 2 3
1 6 7
2 10 11
Point to note here is we are operating on columns
Similarly, if your operation requires traversing from top to bottom/bottom to top in a dataframe, you are merging rows. This is axis=0.

I will explicitly avoid using 'row-wise' or 'along the columns', since people may interpret them in exactly the wrong way.
Analogy first. Intuitively, you would expect that pandas.DataFrame.drop(axis='column') drops a column from N columns and gives you (N - 1) columns. So you can pay NO attention to rows for now (and remove word 'row' from your English dictionary.) Vice versa, drop(axis='row') works on rows.
In the same way, sum(axis='column') works on multiple columns and gives you 1 column. Similarly, sum(axis='row') results in 1 row. This is consistent with its simplest form of definition, reducing a list of numbers to a single number.
In general, with axis=column, you see columns, work on columns, and get columns. Forget rows.
With axis=row, change perspective and work on rows.
0 and 1 are just aliases for 'row' and 'column'. It's the convention of matrix indexing.

I have been trying to figure out the axis for the last hour as well. The language in all the above answers, and also the documentation is not at all helpful.
To answer the question as I understand it now, in Pandas, axis = 1 or 0 means which axis headers do you want to keep constant when applying the function.
Note: When I say headers, I mean index names
Expanding your example:
+------------+---------+--------+
| | A | B |
+------------+---------+---------
| X | 0.626386| 1.52325|
+------------+---------+--------+
| Y | 0.626386| 1.52325|
+------------+---------+--------+
For axis=1=columns : We keep columns headers constant and apply the mean function by changing data.
To demonstrate, we keep the columns headers constant as:
+------------+---------+--------+
| | A | B |
Now we populate one set of A and B values and then find the mean
| | 0.626386| 1.52325|
Then we populate next set of A and B values and find the mean
| | 0.626386| 1.52325|
Similarly, for axis=rows, we keep row headers constant, and keep changing the data:
To demonstrate, first fix the row headers:
+------------+
| X |
+------------+
| Y |
+------------+
Now populate first set of X and Y values and then find the mean
+------------+---------+
| X | 0.626386
+------------+---------+
| Y | 0.626386
+------------+---------+
Then populate the next set of X and Y values and then find the mean:
+------------+---------+
| X | 1.52325 |
+------------+---------+
| Y | 1.52325 |
+------------+---------+
In summary,
When axis=columns, you fix the column headers and change data, which will come from the different rows.
When axis=rows, you fix the row headers and change data, which will come from the different columns.

Say for example, if you use df.shape then you will get a tuple containing the number of rows & columns in the data frame as the output.
In [10]: movies_df.shape
Out[10]: (1000, 11)
In the example above, there are 1000 rows & 11 columns in the movies data frame where 'row' is mentioned in the index 0 position & 'column' in the index 1 position of the tuple. Hence 'axis=1' denotes column & 'axis=0' denotes row.
Credits: Github

On important thing to keep in mind it that when you are using functions such as mean, median etc., you are basically doing numpy aggregation. Think of aggregation as getting a final single output, which can be column wise, row wise or a single number for the entire dataset.
So when we say aggregation in arrays, say numpy.sum(data, axis = 0), what we really mean is that we want to remove that particular axis (here 0 axis).
Example: For this particular dataset if we compute the sum by axis = 0, we are actually interested in removing (aggregating) the zero axis. Once we remove the zero axis, the aggregation along zero axis will lead to [1,4,3] being equal to 8, [2,3,6] to 11 and [5,7,9] to 21. Similar logic can be extended to axis = 1.
In case of drop, concat and some other functions, we are actually not
aggregating the results.
The mental model I use for intuition:
Think that we have placed a Kangaroo/Frog in each of the cells in the first column when axis = 0 or along the first row if axis = 1.
Case: When axis = zero
Think of plus green shape as a frog.
Axis zero means movement along the rows
Sum: Say we are computing sum, then first they will compute the sum of their positions (r1c1, r2c1, r3c1) [1,4,3] = [8]. Then their next move will also be along the row as axis = 0. Their new positions are in the next pic (below).
Drop: If along the row they encounter any NaN in (r1c1, r2c1, r3c1), they will remove the corresponding row as axis = 0
Sum: Now again, they will compute the sum of their positions (r1c2, r2c2, r3c2) [2,3,6] = [11] and similarly they will move one step forward along the row and compute the sum for the third column [21].
Drop: If along the row they encounter any NaN in (r1c2, r2c2, r3c2), they will remove the corresponding row as axis = 0. Similar logic can be extended for different axis and additional rows/columns.

I think there is an another way to understand it.
For a np.array,if we want eliminate columns we use axis = 1; if we want eliminate rows, we use axis = 0.
np.mean(np.array(np.ones(shape=(3,5,10))),axis = 0).shape # (5,10)
np.mean(np.array(np.ones(shape=(3,5,10))),axis = 1).shape # (3,10)
np.mean(np.array(np.ones(shape=(3,5,10))),axis = (0,1)).shape # (10,)
For pandas object, axis = 0 stands for row-wise operation and axis = 1 stands for column-wise operation. This is different from numpy by definition, we can check definitions from numpy.doc and pandas.doc

Many answers here helped me a lot!
In case you get confused by the different behaviours of axis in Python and MARGIN in R (like in the apply function), you may find a blog post that I wrote of interest: https://accio.github.io/programming/2020/05/19/numpy-pandas-axis.html.
In essence:
Their behaviours are, intriguingly, easier to understand with three-dimensional array than with two-dimensional arrays.
In Python packages numpy and pandas, the axis parameter in sum actually specifies numpy to calculate the mean of all values that can be fetched in the form of array[0, 0, ..., i, ..., 0] where i iterates through all possible values. The process is repeated with the position of i fixed and the indices of other dimensions vary one after the other (from the most far-right element). The result is a n-1-dimensional array.
In R, the MARGINS parameter let the apply function calculate the mean of all values that can be fetched in the form of array[, ... , i, ... ,] where i iterates through all possible values. The process is not repeated when all i values have been iterated. Therefore, the result is a simple vector.

Arrays are designed with so-called axis=0 and rows positioned vertically versus axis=1 and columns positioned horizontally. Axis refers to the dimension of the array.

logically adding in numpy

I want to do the following operation. But It likes the histogram operation.
maxIndex = 6
dst =zeros((1,6))
a =array([1,2,3,4,7,0,3,4,5,7])
index=array([1,1,1,3,3,4,4,5,5,5])
a's length == index's length,
for i in (a.size):
dst[index[i]] = dst[index[i]] + a[i]
How can I do this more pythonic. and more efficiently

If I understand correctly, I think you are looking for numpy.bincount:
dst = numpy.bincount(index, weights=a, minlength=maxIndex)
This give me array([ 0., 6., 0., 11., 3., 16.]) as the output. If you don't want to calculate maxIndex by hand, you can omit minlength parameter from the function call and numpy will return an appropriately-sized array for you.

expanding (adding a row or column) a scipy.sparse matrix

Suppose I have a NxN matrix M (lil_matrix or csr_matrix) from scipy.sparse, and I want to make it (N+1)xN where M_modified[i,j] = M[i,j] for 0 <= i < N (and all j) and M[N,j] = 0 for all j. Basically, I want to add a row of zeros to the bottom of M and preserve the remainder of the matrix. Is there a way to do this without copying the data?

Scipy doesn't have a way to do this without copying the data but you can do it yourself by changing the attributes that define the sparse matrix.
There are 4 attributes that make up the csr_matrix:
data: An array containing the actual values in the matrix
indices: An array containing the column index corresponding to each value in data
indptr: An array that specifies the index before the first value in data for each row. If the row is empty then the index is the same as the previous column.
shape: A tuple containing the shape of the matrix
If you are simply adding a row of zeros to the bottom all you have to do is change the shape and indptr for your matrix.
x = np.ones((3,5))
x = csr_matrix(x)
x.toarray()
>> array([[ 1., 1., 1., 1., 1.],
[ 1., 1., 1., 1., 1.],
[ 1., 1., 1., 1., 1.]])
# reshape is not implemented for csr_matrix but you can cheat and do it yourself.
x._shape = (4,5)
# Update indptr to let it know we added a row with nothing in it. So just append the last
# value in indptr to the end.
# note that you are still copying the indptr array
x.indptr = np.hstack((x.indptr,x.indptr[-1]))
x.toarray()
array([[ 1., 1., 1., 1., 1.],
[ 1., 1., 1., 1., 1.],
[ 1., 1., 1., 1., 1.],
[ 0., 0., 0., 0., 0.]])
Here is a function to handle the more general case of vstacking any 2 csr_matrices. You still end up copying the underlying numpy arrays but it is still significantly faster than the scipy vstack method.
def csr_vappend(a,b):
""" Takes in 2 csr_matrices and appends the second one to the bottom of the first one.
Much faster than scipy.sparse.vstack but assumes the type to be csr and overwrites
the first matrix instead of copying it. The data, indices, and indptr still get copied."""
a.data = np.hstack((a.data,b.data))
a.indices = np.hstack((a.indices,b.indices))
a.indptr = np.hstack((a.indptr,(b.indptr + a.nnz)[1:]))
a._shape = (a.shape[0]+b.shape[0],b.shape[1])
return a

Not sure if you're still looking for a solution, but maybe others can look into hstack and vstack - http://docs.scipy.org/doc/scipy/reference/generated/scipy.sparse.hstack.html. I think we can define a csr_matrix for the single additional row and then vstack it with the previous matrix.

I don't think that there is any way to really escape from doing the copying. Both of those types of sparse matrices store their data as Numpy arrays (in the data and indices attributes for csr and in the data and rows attributes for lil) internally and Numpy arrays can't be extended.
Update with more information:
LIL does stand for LInked List, but the current implementation doesn't quite live up to the name. The Numpy arrays used for data and rows are both of type object. Each of the objects in these arrays are actually Python lists (an empty list when all values are zero in a row). Python lists aren't exactly linked lists, but they are kind of close and quite frankly a better choice due to O(1) look-up. Personally, I don't immediately see the point of using a Numpy array of objects here rather than just a Python list. You could fairly easily change the current lil implementation to use Python lists instead which would allow you to add a row without copying the whole matrix.