Using the .loc function of the pandas dataframe

You can use between to get Boolean values, then astype to convert from Boolean values to 0/1:

dtaframe['b'] = dtaframe['a'].between(0, 5, inclusive=False).astype(int)

The resulting output:

   a  b
0  1  1
1  0  0
2  1  1
3  0  0
4  1  1
5  3  1
6  4  1
7  6  0
8  4  1
9  6  0

Edit

For multiple ranges, you could use pandas.cut:

dtaframe['b'] = pd.cut(dtaframe['a'], bins=[0,1,6,9], labels=False, include_lowest=True)

You'll need to be careful about how you define bins. Using labels=False will return integer indicators for each bin, which happens to correspond with the labels you provided. You could also manually specify the labels for each bin, e.g. labels=[0,1,2], labels=[0,17,19], labels=['a','b','c'], etc. You may need to use astype if you manually specify the labels, as they'll be returned as categories.

Alternatively, you could combine loc and between to manually specify each range:

dtaframe.loc[dtaframe['a'].between(0,1), 'b'] = 0
dtaframe.loc[dtaframe['a'].between(2,6), 'b'] = 1
dtaframe.loc[dtaframe['a'].between(7,9), 'b'] = 2

Try this with np.where:

dtaframe['b'] = np.where(([dtaframe['a'] > 4) | (dtaframe['a']==0),0, dtaframe['a'])

When using comparison operators and boolean logic to filter dataframes you can't use the pythonic idiom of a < myseries < b. Instead you need to (a < myseries) & (myseries < b)

cond1 = (0 < dtaframe['a'])
cond2 = (dtaframe['a'] <= 5)
dtaframe['b'] = (cond1 & cond2) * 1