Granted, the answer I linked in the comments is not very helpful. You can specify your own string converter like so.

In [25]: pd.set_option('display.float_format', lambda x: '%.3f' % x)

In [28]: Series(np.random.randn(3))*1000000000
Out[28]: 
0    -757322420.605
1   -1436160588.997
2   -1235116117.064
dtype: float64

I'm not sure if that's the preferred way to do this, but it works.

Converting numbers to strings purely for aesthetic purposes seems like a bad idea, but if you have a good reason, this is one way:

In [6]: Series(np.random.randn(3)).apply(lambda x: '%.3f' % x)
Out[6]: 
0     0.026
1    -0.482
2    -0.694
dtype: object

This worked for me, worth giving a try:

df['col_name'] = df['col_name'].astype('int64') 

You can use map or apply, as mentioned in this comment:

print (df.userid.map(lambda x: '{:.0f}'.format(x)))
0             nan
1    109117800000
2    113785600000
Name: userid, dtype: object

df.userid = df.userid.map(lambda x: '{:.0f}'.format(x))
print (df)
         userid
0           nan
1  109117800000
2  113785600000

I wondered whether map would be faster, but it is the same:

#[300000 rows x 1 columns]
df = pd.concat([df]*100000).reset_index(drop=True)
#print (df)

In [40]: %timeit (df.userid.map(lambda x: '{:.0f}'.format(x)))
1 loop, best of 3: 211 ms per loop

In [41]: %timeit (df.userid.apply(lambda x: '{:.0f}'.format(x)))
1 loop, best of 3: 210 ms per loop

Another solution is to_string, but it is slow:

print(df.userid.to_string(float_format='{:.0f}'.format))
0            nan
1   109117800000
2   113785600000

In [41]: (df.userid.to_string(float_format='{:.0f}'.format))
1 loop, best of 3: 2.52 s per loop

OK, I figured this out, you can use set_option and pass a format string to option 'display.float_format':

In [76]:
pd.set_option('display.float_format', '{:.2g}'.format)

In [78]:
pd.Series(data=[0.00000001])

Out[78]:
0   1e-08
dtype: float64

EDIT

to match your desired output:

In [79]:
pd.set_option('display.float_format', '{:.2E}'.format)
pd.Series(data=[0.00000001])

Out[79]:
0   1.00E-08
dtype: float64

I realised it was the infinity statement causing the issue in my data. Removing this with a find and replace worked.

@Anton Protopopov answer also works as did @DSM's comment regarding me not typing df['speed'] = df['speed'].astype(float).

Thanks for the help.

you can use float_format parameter when calling .to_csv() function:

In [207]: df
Out[207]:
   co_A          co_B  co_C
0   167  0.000000e+00  59.6
1   168  0.000000e+00  60.6
2   191  8.000000e-09  72.6
3   197 -4.771800e-06  12.3
4   197  0.000000e+00  92.4
5   198  0.000000e+00  39.5

In [208]: fn = r'D:\temp\.data\out.csv'

In [209]: df.to_csv(fn, index=False, sep='\t', float_format='%.6f')

out.csv:

co_A    co_B    co_C
167 0.000000    59.600000
168 0.000000    60.600000
191 0.000000    72.600000
197 -0.000005   12.300000
197 0.000000    92.400000
198 0.000000    39.500000