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Difference between df.loc['col name'], df.loc[index]['col name'] and df.loc[index, 'col name'] in pandas?

I have a dataframe df with a column name 'Store'. If I want to retrieve the column, the following lines work equally well - df['Store'] or df[:]['Store'] or df[:,'Store'].
What is the difference between the two? And should one be used over the other?
Thank you.

df.loc[index, 'col name'] is more idiomatic and preferred, especially if you want to filter rows
Demo: for 1.000.000 x 3 shape DF
In [26]: df = pd.DataFrame(np.random.rand(10**6,3), columns=list('abc'))
In [27]: %timeit df[df.a < 0.5]['a']
10 loops, best of 3: 45.8 ms per loop
In [28]: %timeit df.loc[df.a < 0.5]['a']
10 loops, best of 3: 45.8 ms per loop
In [29]: %timeit df.loc[df.a < 0.5, 'a']
10 loops, best of 3: 37 ms per loop
For construction where you need only one column and don't filter rows like df[:]['Store'] - it's better to use simply df['Store']:
In [30]: %timeit df[:]['a']
1000 loops, best of 3: 436 µs per loop
In [31]: %timeit df.loc[:]['a']
10000 loops, best of 3: 25.9 µs per loop
In [36]: %timeit df['a'].loc[:]
10000 loops, best of 3: 26.5 µs per loop
In [32]: %timeit df.loc[:, 'a']
10000 loops, best of 3: 126 µs per loop
In [33]: %timeit df['a']
The slowest run took 5.08 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 8.17 µs per loop
Unconditional access of multiple columns:
In [34]: %timeit df[['a','b']]
10 loops, best of 3: 22 ms per loop
In [35]: %timeit df.loc[:, ['a','b']]
10 loops, best of 3: 22.6 ms per loop

Fast advanced indexing in numpy

I'm trying to take a slice from a large numpy array as quickly as possible using fancy indexing. I would be happy returning a view, but advanced indexing returns a copy.
I've tried solutions from here and here with no joy so far.
Toy data:
data = np.random.randn(int(1e6), 50)
keep = np.random.rand(len(data))>0.5
Using the default method:
%timeit data[keep]
10 loops, best of 3: 86.5 ms per loop
Numpy take:
%timeit data.take(np.where(keep)[0], axis=0)
%timeit np.take(data, np.where(keep)[0], axis=0)
10 loops, best of 3: 83.1 ms per loop
10 loops, best of 3: 80.4 ms per loop
Method from here:
rows = np.where(keep)[0]
cols = np.arange(a.shape[1])
%timeit (a.ravel()[(cols + (rows * a.shape[1]).reshape((-1,1))).ravel()]).reshape(rows.size, cols.size)
10 loops, best of 3: 159 ms per loop
Whereas if you're taking a view of the same size:
%timeit data[1:-1:2, :]
1000000 loops, best of 3: 243 ns per loop

There's no way to do this with a view. A view needs consistent strides, while your data is randomly scattered throughout the original array.

Fast replacement of tzinfo of a pandas.Series of datetime

I have a pandas.Seriesof datetime and need to replace the tzinfo for every element in it.
I know how to do it using apply with python function but it's very slow: ~16s for 1M elements on a MacBookPro
In [71]: s = pd.date_range('2015-1-1', freq='h', periods=1e6).to_series().reset_index(drop=True)
In [72]: %timeit s.apply(lambda x: x.replace(tzinfo=pytz.utc))
1 loops, best of 3: 16.7 s per loop
Is there a numpy ufunc function for it?

Use dt.localize:
In [33]:
import pytz
%timeit s.dt.tz_localize(pytz.utc)
%timeit s.apply(lambda x: x.replace(tzinfo=pytz.utc))
10 loops, best of 3: 107 ms per loop
1 loops, best of 3: 10.4 s per loop
As you can see ~100X faster

python pandas: why map is faster?

in pandas' manual, there is this example about indexing:
In [653]: criterion = df2['a'].map(lambda x: x.startswith('t'))
In [654]: df2[criterion]
then Wes wrote:
**# equivalent but slower**
In [655]: df2[[x.startswith('t') for x in df2['a']]]
can anyone here explain a bit why the map approach is faster? Is this a python feature or this is a pandas feature?

Arguments about why a certain way of doing things in Python "should be" faster can't be taken too seriously, because you're often measuring implementation details which may behave differently in certain situations. As a result, when people guess what should be faster, they're often (usually?) wrong. For example, I find that map can actually be slower. Using this setup code:
import numpy as np, pandas as pd
import random, string
def make_test(num, width):
s = [''.join(random.sample(string.ascii_lowercase, width)) for i in range(num)]
df = pd.DataFrame({"a": s})
return df
Let's compare the time they take to make the indexing object -- whether a Series or a list -- and the resulting time it takes to use that object to index into the DataFrame. It could be, for example, that making a list is fast but before using it as an index it needs to be internally converted to a Series or an ndarray or something and so there's extra time added there.
First, for a small frame:
>>> df = make_test(10, 10)
>>> %timeit df['a'].map(lambda x: x.startswith('t'))
10000 loops, best of 3: 85.8 µs per loop
>>> %timeit [x.startswith('t') for x in df['a']]
100000 loops, best of 3: 15.6 µs per loop
>>> %timeit df['a'].str.startswith("t")
10000 loops, best of 3: 118 µs per loop
>>> %timeit df[df['a'].map(lambda x: x.startswith('t'))]
1000 loops, best of 3: 304 µs per loop
>>> %timeit df[[x.startswith('t') for x in df['a']]]
10000 loops, best of 3: 194 µs per loop
>>> %timeit df[df['a'].str.startswith("t")]
1000 loops, best of 3: 348 µs per loop
and in this case the listcomp is fastest. That doesn't actually surprise me too much, to be honest, because going via a lambda is likely to be slower than using str.startswith directly, but it's really hard to guess. 10 is small enough we're probably still measuring things like setup costs for Series; what happens in a larger frame?
>>> df = make_test(10**5, 10)
>>> %timeit df['a'].map(lambda x: x.startswith('t'))
10 loops, best of 3: 46.6 ms per loop
>>> %timeit [x.startswith('t') for x in df['a']]
10 loops, best of 3: 27.8 ms per loop
>>> %timeit df['a'].str.startswith("t")
10 loops, best of 3: 48.5 ms per loop
>>> %timeit df[df['a'].map(lambda x: x.startswith('t'))]
10 loops, best of 3: 47.1 ms per loop
>>> %timeit df[[x.startswith('t') for x in df['a']]]
10 loops, best of 3: 52.8 ms per loop
>>> %timeit df[df['a'].str.startswith("t")]
10 loops, best of 3: 49.6 ms per loop
And now it seems like the map is winning when used as an index, although the difference is marginal. But not so fast: what if we manually turn the listcomp into an array or a Series?
>>> %timeit df[np.array([x.startswith('t') for x in df['a']])]
10 loops, best of 3: 40.7 ms per loop
>>> %timeit df[pd.Series([x.startswith('t') for x in df['a']])]
10 loops, best of 3: 37.5 ms per loop
and now the listcomp wins again!
Conclusion: who knows? But never believe anything without timeit results, and even then you have to ask whether you're testing what you think you are.

Why is numpy.array() is sometimes very slow?

I'm using the numpy.array() function to create numpy.float64 ndarrays from lists.
I noticed that this is very slow when either the list contains None or a list of lists is provided.
Below are some examples with times. There are obvious workarounds but why is this so slow?
Examples for list of None:
### Very slow to call array() with list of None
In [3]: %timeit numpy.array([None]*100000, dtype=numpy.float64)
1 loops, best of 3: 240 ms per loop
### Problem doesn't exist with array of zeroes
In [4]: %timeit numpy.array([0.0]*100000, dtype=numpy.float64)
100 loops, best of 3: 9.94 ms per loop
### Also fast if we use dtype=object and convert to float64
In [5]: %timeit numpy.array([None]*100000, dtype=numpy.object).astype(numpy.float64)
100 loops, best of 3: 4.92 ms per loop
### Also fast if we use fromiter() insead of array()
In [6]: %timeit numpy.fromiter([None]*100000, dtype=numpy.float64)
100 loops, best of 3: 3.29 ms per loop
Examples for list of lists:
### Very slow to create column matrix
In [7]: %timeit numpy.array([[0.0]]*100000, dtype=numpy.float64)
1 loops, best of 3: 353 ms per loop
### No problem to create column vector and reshape
In [8]: %timeit numpy.array([0.0]*100000, dtype=numpy.float64).reshape((-1,1))
100 loops, best of 3: 10 ms per loop
### Can use itertools to flatten input lists
In [9]: %timeit numpy.fromiter(itertools.chain.from_iterable([[0.0]]*100000),dtype=numpy.float64).reshape((-1,1))
100 loops, best of 3: 9.65 ms per loop

I've reported this as a numpy issue. The report and patch files are here:
https://github.com/numpy/numpy/issues/3392
After patching:
# was 240 ms, best alternate version was 3.29
In [5]: %timeit numpy.array([None]*100000)
100 loops, best of 3: 7.49 ms per loop
# was 353 ms, best alternate version was 9.65
In [6]: %timeit numpy.array([[0.0]]*100000)
10 loops, best of 3: 23.7 ms per loop

My guess would be that the code for converting lists just calls float on everything. If the argument defines __float__, we call that, otherwise we treat it like a string (throwing an exception on None, we catch that and puts in np.nan). The exception handling should be relatively slower.
Timing seems to verify this hypothesis:
import numpy as np
%timeit [None] * 100000
> 1000 loops, best of 3: 1.04 ms per loop
%timeit np.array([0.0] * 100000)
> 10 loops, best of 3: 21.3 ms per loop
%timeit [i.__float__() for i in [0.0] * 100000]
> 10 loops, best of 3: 32 ms per loop
def flt(d):
try:
return float(d)
except:
return np.nan
%timeit np.array([None] * 100000, dtype=np.float64)
> 1 loops, best of 3: 477 ms per loop
%timeit [flt(d) for d in [None] * 100000]
> 1 loops, best of 3: 328 ms per loop
Adding another case just to be obvious about where I'm going with this. If there was an explicit check for None, it would not be this slow above:
def flt2(d):
if d is None:
return np.nan
try:
return float(d)
except:
return np.nan
%timeit [flt2(d) for d in [None] * 100000]
> 10 loops, best of 3: 45 ms per loop