There is a pandas dataframe:
df = pd.DataFrame({'c1':['a','b','c','d'],'c2':[1,2,3,4]})
c1 c2
0 a 1
1 b 2
2 c 3
3 d 4
And a pandas Series:
list1 = pd.Series(['b','c','e','f'])
Out[6]:
0 a
1 b
2 c
3 e
How to create a new data frame that contains rows where c1 is in list1.
output:
c1 c2
0 b 2
1 c 3
You can use df.isin:
In [582]: df[df.c1.isin(list1)]
Out[582]:
c1 c2
1 b 2
2 c 3
Or, using df.loc, if you want to modify your slice:
In [584]: df.loc[df.c1.isin(list1), :]
Out[584]:
c1 c2
1 b 2
2 c 3
Using query
In [1133]: df.query('c1 in @list1')
Out[1133]:
c1 c2
1 b 2
2 c 3
Or, using isin
In [1134]: df[df.c1.isin(list1)]
Out[1134]:
c1 c2
1 b 2
2 c 3
Both @JohnGalt's and @COLDSPEED's answers are more idiomatic pandas. Please don't use these answers. They are intended to be fun and illustrative of other parts of the pandas and numpy api.
Alt 1
This is utilizing numpy.in1d which acts as a proxy for pd.Series.isin
df[np.in1d(df.c1.values, list1.values)]
c1 c2
1 b 2
2 c 3
Alt 2
Use set logic
df[df.c1.apply(set) & set(list1)]
c1 c2
1 b 2
2 c 3
Alt 3
Use pd.Series.str.match
df[df.c1.str.match('|'.join(list1))]
c1 c2
1 b 2
2 c 3
For the sake of completenes
yet another way (definitely not the best one) to achieve that:
In [4]: df.merge(list1.to_frame(name='c1'))
Out[4]:
c1 c2
0 b 2
1 c 3