What you see is just the default string representation with 6 decimal places. You can set you own display format option with

pd.options.display.float_format = '{:.10f}'.format

to show 10 places. Alternatively you can confirm the number if you look at the result of say df.loc[1,'lng'].

To set the option only temporarily you can use an option context:

with pd.option_context('display.float_format', '{:.10f}'.format):
    print(df)

NOTE: pd.convert_objects has now been deprecated. You should use pd.Series.astype(float) or pd.to_numeric as described in other answers.

This is available in 0.11. Forces conversion (or set's to nan) This will work even when astype will fail; its also series by series so it won't convert say a complete string column

In [10]: df = DataFrame(dict(A = Series(['1.0','1']), B = Series(['1.0','foo'])))

In [11]: df
Out[11]: 
     A    B
0  1.0  1.0
1    1  foo

In [12]: df.dtypes
Out[12]: 
A    object
B    object
dtype: object

In [13]: df.convert_objects(convert_numeric=True)
Out[13]: 
   A   B
0  1   1
1  1 NaN

In [14]: df.convert_objects(convert_numeric=True).dtypes
Out[14]: 
A    float64
B    float64
dtype: object

You can convert most of the columns by just calling convert_objects:

In [36]:

df = df.convert_objects(convert_numeric=True)
df.dtypes
Out[36]:
Date         object
WD            int64
Manpower    float64
2nd          object
CTR          object
2ndU        float64
T1            int64
T2          int64
T3           int64
T4        float64
dtype: object

For column '2nd' and 'CTR' we can call the vectorised str methods to replace the thousands separator and remove the '%' sign and then astype to convert:

In [39]:

df['2nd'] = df['2nd'].str.replace(',','').astype(int)
df['CTR'] = df['CTR'].str.replace('%','').astype(np.float64)
df.dtypes
Out[39]:
Date         object
WD            int64
Manpower    float64
2nd           int32
CTR         float64
2ndU        float64
T1            int64
T2            int64
T3            int64
T4           object
dtype: object
In [40]:

df.head()
Out[40]:
        Date  WD  Manpower   2nd   CTR  2ndU   T1  　　T2   T3     T4
0   2013/4/6   6       NaN  2645  5.27  0.29  407   533  454    368
1   2013/4/7   7       NaN  2118  5.89  0.31  257   659  583    369
2  2013/4/13   6       NaN  2470  5.38  0.29  354   531  473  　　383
3  2013/4/14   7       NaN  2033  6.77  0.37  396   748  681    458
4  2013/4/20   6       NaN  2690  5.38  0.29  361   528  541    381

Or you can do the string handling operations above without the call to astype and then call convert_objects to convert everything in one go.

UPDATE

Since version 0.17.0 convert_objects is deprecated and there isn't a top-level function to do this so you need to do:

df.apply(lambda col:pd.to_numeric(col, errors='coerce'))

See the docs and this related question: pandas: to_numeric for multiple columns

If memory is a problem, and if you know the size of the field ahead of time, you probably don't want to read the entire file in the first place. Something like this is probably more appropriate:

#allocate memory (np.empty would work too and be marginally faster, 
#                 but probably not worth mentioning).
a=np.zeros((3000,300),dtype=np.float32)  
with open(filename) as f:
    for i,line in enumerate(f):
        a[i,:]=map(np.float32,line.split()) 

from a couple quick (and surprising) tests on my machine, it appears that the map may not even be necessary:

a=np.zeros((3000,300),dtype=np.float32)  
with open(filename) as f:
    for i,line in enumerate(f):
        a[i,:]=line.split() 

This might not be the fastest, but certainly it'll be the most memory efficient way to do it.

Some tests:

import numpy as np

def func1():   #No map -- And pretty speedy :-).
    a=np.zeros((3000,300),dtype=np.float32)
    with open('junk.txt') as f:
        for i,line in enumerate(f):
            a[i,:]=line.split()

def func2():
    a=np.zeros((3000,300),dtype=np.float32)
    with open('junk.txt') as f:
        for i,line in enumerate(f):
            a[i,:]=map(np.float32,line.split())

def func3():
    a=np.zeros((3000,300),dtype=np.float32)
    with open('junk.txt') as f:
        for i,line in enumerate(f):
            a[i,:]=map(float,line.split())

import timeit

print timeit.timeit('func1()',setup='from __main__ import func1',number=3)  #1.36s
print timeit.timeit('func2()',setup='from __main__ import func2',number=3)  #11.53s
print timeit.timeit('func3()',setup='from __main__ import func3',number=3)  #1.72s

With Python < 3 (e.g. 2.6 [see comments] or 2.7), there are two ways to do so.

# Option one
older_method_string = "%.9f" % numvar

# Option two
newer_method_string = "{:.9f}".format(numvar)

But note that for Python versions above 3 (e.g. 3.2 or 3.3), option two is preferred.

For more information on option two, I suggest this link on string formatting from the Python documentation.

And for more information on option one, this link will suffice and has info on the various flags.

Python 3.6 (officially released in December of 2016), added the f string literal, see more information here, which extends the str.format method (use of curly braces such that f"{numvar:.9f}" solves the original problem), that is,

# Option 3 (versions 3.6 and higher)
newest_method_string = f"{numvar:.9f}"

solves the problem. Check out @Or-Duan's answer for more info, but this method is fast.

Well, if you're reading the data in as a list, just do np.array(map(float, list_of_strings)) (or equivalently, use a list comprehension). (In Python 3, you'll need to call list on the map return value if you use map, since map returns an iterator now.)

However, if it's already a numpy array of strings, there's a better way. Use astype().

import numpy as np
x = np.array(['1.1', '2.2', '3.3'])
y = x.astype(np.float)

>>> a = "545.2222"
>>> float(a)
545.22220000000004
>>> int(float(a))
545