You have a few options...

1) convert everything to integers.

df.astype(int)
          <=35  >35
Cut-off            
Calcium      0    1
Copper       1    0
Helium       0    8
Hydrogen     0    1

2) Use round:

>>> df.round()
          <=35  >35
Cut-off            
Calcium      0    1
Copper       1    0
Helium       0    8
Hydrogen     0    1

but not always great...

>>> (df - .2).round()
          <=35  >35
Cut-off            
Calcium     -0    1
Copper       1   -0
Helium      -0    8
Hydrogen    -0    1

3) Change your display precision option in Pandas.

pd.set_option('precision', 0)

>>> df
          <=35  >35
Cut-off            
Calcium      0    1
Copper       1    0
Helium       0    8
Hydrogen     0    1 

You need to re-assign dataframe (which is, what I suppose your error is):

>>> import pandas as pd
>>> df = pd.DataFrame(data={"col": [24.00, 2.00, 3.00]})
>>> df.dtypes
col    float64
dtype: object
>>> df
    col
0  24.0
1   2.0
2   3.0
>>> df=df.astype(int)
>>> df
   col
0   24
1    2
2    3
>>> df.dtypes
col    int32
dtype: object

The following would preserve all digits and just get rid of the final decimal point which is what you wanted:

In [80]:
t="""val
1.2345.678"""
df = pd.read_csv(io.StringIO(t))
df

Out[80]:
          val
0  1.2345.678

In [94]:    
(df['val'].str[0:df['val'].str.rfind('.')[0]]+df['val'].str.split('.').str[-1]).astype(np.float64)

Out[94]:
0    1.234568
dtype: float64

Note that the above shows a display truncation, the full value is present

So the above slices your string from beginning to the position of the last decimal point, I then split the string and add the last split, you can then convert to a float using astype

EDIT

A better way I think is for the second part not to split but rather re-use rfind for the positioning:

In [113]:
df['val'].str[0:df['val'].str.rfind('.')[0]]+df['val'].str[df['val'].str.rfind('.')[0]+1:]

Out[113]:
0    1.2345678
Name: val, dtype: object

Use a function and apply to whole column:

In [94]:

df = pd.DataFrame({'Movies':['Save the last dance', '2012.0']})
df
Out[94]:
                Movies
0  Save the last dance
1               2012.0

[2 rows x 1 columns]

In [95]:

def trim_fraction(text):
    if '.0' in text:
        return text[:text.rfind('.0')]
    return text

df.Movies = df.Movies.apply(trim_fraction)

In [96]:

df
Out[96]:
                Movies
0  Save the last dance
1                 2012

[2 rows x 1 columns]

You can try using as df['col'] = (df['col']*100).astype(int) as below:

df = pd.DataFrame({'col': [1.10, 2.20, 3.30, 4.40]})
df['col'] = (df['col']*100).astype(int)
print(df)

Output:

   col
0  110
1  220
2  330
3  440

You can call int() on the end result:

>>> int(2.0)
2

You can convert the type with .astype

In [312]: df.columns = df.columns.astype(int)

In [313]: df
Out[313]:
  2006 2007 2008 2009
0  foo  foo  bar  bar
1  foo  foo  bar  bar

Or use .map and convert to string type.

In [338]: df.columns.map('{:g}'.format)
Out[338]: Index(['2006', '2007', '2008', '2009'], dtype='object')

In [319]: df.columns.map(int)
Out[319]: Int64Index([2006, 2007, 2008, 2009], dtype='int64')

This works:

df['energyConsumptionInKWh'] = (df['energyConsumptionInKWh'] * 1000).astype(int).astype(str).str.replace('.','')