Slice Pandas dataframe by index values that are (not) in a list

Use isin on the index and invert the boolean index to perform label selection:

In [239]:

df = pd.DataFrame({'a':np.random.randn(5)})
df
Out[239]:
          a
0 -0.548275
1 -0.411741
2 -1.187369
3  1.028967
4 -2.755030
In [240]:

t = [2,4]
df.loc[~df.index.isin(t)]
Out[240]:
          a
0 -0.548275
1 -0.411741
3  1.028967

You could use set() to create the difference between your original indices and those that you want to remove:

df.loc[set(df.index) - set(blacklist)]

It has the advantage of being parsimonious, as well as being easier to read than a list comprehension.

Thanks to ASGM; I found that I needed to turn the set into a list to make it work with a MultiIndex:

mi1 = pd.MultiIndex.from_tuples([("a", 1), ("a", 2), ("b", 1), ("b", 2)])
df1 = pd.DataFrame(data={"aaa":[1,2,3,4]}, index=mi1)
setValid = set(df1.index) - set([("a", 2)])
df1.loc[list(setValid)] # works
df1.loc[setValid] # fails

(sorry can't comment, insufficient rep)

If you are looking for a way to select all rows that are outside a condition you can use np.invert() given that the condition returns an array of booleans.

df.loc[np.invert(({condition 1}) & (condition 2))]

import pandas as pd
df = pd.DataFrame(data=[5,6,7,8], index=[1,2,3,4], columns=['D',])
blacklist = [2,3]
#your current way ...
ix=[i for i in df.index if i not in blacklist]  
df_select=df.loc[ix]

# use a mask
mask = [True if x else False for x in df.index if x not in blacklist]
df.loc[mask]

http://pandas.pydata.org/pandas-docs/dev/indexing.html#indexing-label actually, loc and iloc both take a boolean array, in this case the mask. from now on you can reuse this mask and should be more efficient.