Numpy slicing allows you to input a list of indices to an array so that you can slice to the exact values you want.

For example:

    import numpy as np
    a = np.random.randn(10)
    a[[2,4,6,8]]

This will return the 2nd, 4th, 6th, and 8th array elements (keeping in mind that python indices start from 0). So, if you want every 2nd element starting from an index x, you can simply populate a list with those elements and then feed that list into the array to get the elements you want, e.g:

    idx = list(range(2,10,2))
    a[idx]

This again returns the desired elements (index 2,4,6,8).

for i in range(15):
    print i #will print out 0..14

for i in range(1, 15):
    print i # will print out 1..14


for i in range (a, b, s):
    print i # will print a..b-1 counting by s. interestingly if while counting by the step 's' you exceed b, it will stop at the last 'reachable' number, example

for i in range(1, 10, 3):
    print i

> 1
> 4
> 7

List Splicing:

a = "hello" # there are 5 characters, so the characters are accessible on indexes 0..4

a[1] = 'e'
a[1:2] = 'e' # because the number after the colon is not reached.

a[x:y] = all characters starting from the character AT index 'x' and ending at the character which is before 'y'

a[x:] = all characters starting from x and to the end of the string

In the future, if you ever wonder what the behavior of python is like, you can try it out in the python shell. just type python in the terminal and you can enter any lines you want (though this is mostly convenient for one-liners rather than scripts).

If you really want to use manual indexing, then dont use enumerate() and just create a range() (or xrange() if Python 2.x) of the right size, ie:

for i in xrange(len(a) - 2):
   # code here

Now you don't have to manually take care of indexes at all - if you want to iterate over (a[x], a[x+1]) pairs all you need is zip():

for x, y in zip(a, a[1:]):
   if abs(x - y) < min:
       min = abs(x - y)

zip(seq1, seq2) will build a list of (seq1[i], seq2[i]) tuples (stopping when the smallest sequence or iterator is exhausted). Using a[1:] as the second sequence, we will have a list of (a[i], a[i+1]) tuples. Then we use tuple unpacking to assign each of the tuple's values to x and y.

But you can also just use the builtin min(iterable) function instead:

min(abs(x - y) for x, y in zip(a, a[1:]))

which is the pythonic way to get the smallest value of any sequence or iterable.

Note that with Python 2.x, if your real list is actually way bigger, you'll benefit from using itertools.izip instead of zip

As as side note, using min (actually using any builtin name) as a variable name is possibly not a good idea as it shadows the builtin in the current namespace. If you get a TypeError: 'int' object is not callable message trying this code you'll know why...

a = [12,11,5,7,2,21,32,13,6,42,1,8,9,0,32,38]
indices = [idx for idx,val in enumerate(a) if val < 10]

This creates a list of indices:

[2, 3, 4, 8, 10, 11, 12, 13]

I would recommend keeping it that way for easy parsing, but you can also turn it into ranges as follows:

ranges = [[]]
for val in indices:
    if not ranges[-1] or ranges[-1][-1] == val-1:
        ranges[-1].append(val)
    else:
        ranges.append([val])

This creates a list of ranges:

[[2, 3, 4], [8], [10, 11, 12, 13]]

Now to take out the middle:

ranges = [[item[0],item[-1]] if len(item) > 1 else item for item in ranges]

Result:

[[2, 4], [8], [10, 13]]

I believe this accomplishes what you want, if I interpreted your question correctly:

indents = ['1', '2', '3', '1', '2', '2', '2', '2', '1', '2']

arraySaved = []; temp = [0] #Initialize temporary list
for idx, i in enumerate(indents):
    if idx==len(indents)-1:
        temp.append(idx)
        arraySaved.append(temp) #Reached end of list
    elif indents[idx+1]<i: #Ending index of temporary list
        temp.append(idx)
        arraySaved.append(temp) #Store temporary list
        temp = []; temp.append(idx+1) #Reset temporary list and begin new one

print(arraySaved)

Yields:

[[0, 2], [3, 7], [8, 9]]

Keep in mind that your desired output is the upper and lower bounds of the row indices after being separated into individually increasing indent counts. Therefore, you do not actually need the list rows, since you can just enumerate the list indents. The answer above is equivalent to your desired output if you keep in mind that Python indexes from 0, not 1.

Figured I would add that if you really want the row numbers indexed from 1, then you can do the following:

arraySaved = [[i+1 for i in j] for j in arraySaved]

Gives:

[[2, 3], [4, 8], [9, 10]]

Explanation

temp is simply a list used to temporarily store the indices of the values in indents that correspond to the starting and ending indices for each individual outputted list eventually stored in arraySaved. We need to initialize temp as well with the first index of the list, i.e. 0.

for idx, i in enumerate(indents): simply loops through the values inside of the list indents, where enumerate also unpacks the indices of the values inside of the list as well

The first if statement accounts for the case that the current index in the loop is the last one in the list, because then idx+1 would exceed the dimensions of the list being iterated on. If either of the criteria in the if statements are satisfied, then the current index is stored in the temp variable. If the ending index criteria is satisfied then the temp list is reset after being appended to arraySaved.

Hope that helps!

Check this out: This returns True if and only if ALL of the numbers in the list n are at least 0 and at most 9, (in range(0, 10))

 all(i in range(0, 10) for i in n)

What condition are you looking to satisfy?

import pandas as pd

df=pd.DataFrame([['This is also a interesting topic',2],['<body> the valley of flowers ...',1],['found in the hilly terrain',5],
             ['we must preserve it </body>',6]],columns=['description','count'])
print(df)
print(len(df[df['count'] != 2].index))

Here, df['count'] != 2 subsets the df, and len(df.index) returns the length of the index.

Updated; note that I used str.contains(), rather than explicitly looking for starting or ending strings.

df2 = df[(df.description.str.contains('<body>') | (df.description.str.contains('</body>')))]
print(df2)
print(len(df2.index))

help from: Check if string is in a pandas dataframe