- Python Pandas Tutorial
- Python Pandas - Home
- Python Pandas - Introduction
- Python Pandas - Environment Setup
- Introduction to Data Structures
- Python Pandas - Series
- Python Pandas - DataFrame
- Python Pandas - Panel
- Python Pandas - Basic Functionality
- Descriptive Statistics
- Function Application
- Python Pandas - Reindexing
- Python Pandas - Iteration
- Python Pandas - Sorting
- Working with Text Data
- Options & Customization
- Indexing & Selecting Data
- Statistical Functions
- Python Pandas - Window Functions
- Python Pandas - Aggregations
- Python Pandas - Missing Data
- Python Pandas - GroupBy
- Python Pandas - Merging/Joining
- Python Pandas - Concatenation
- Python Pandas - Date Functionality
- Python Pandas - Timedelta
- Python Pandas - Categorical Data
- Python Pandas - Visualization
- Python Pandas - IO Tools
- Python Pandas - Sparse Data
- Python Pandas - Caveats & Gotchas
- Comparison with SQL
- Python Pandas Useful Resources
- Python Pandas - Quick Guide
- Python Pandas - Useful Resources
- Python Pandas - Discussion
Python Pandas - Caveats & Gotchas
Caveats means warning and gotcha means an unseen problem.
Using If/Truth Statement with Pandas
Pandas follows the numpy convention of raising an error when you try to convert something to a bool. This happens in an if or when using the Boolean operations, and, or, or not. It is not clear what the result should be. Should it be True because it is not zerolength? False because there are False values? It is unclear, so instead, Pandas raises a ValueError −
import pandas as pd if pd.Series([False, True, False]): print 'I am True'
Its output is as follows −
ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool() a.item(),a.any() or a.all().
In if condition, it is unclear what to do with it. The error is suggestive of whether to use a None or any of those.
import pandas as pd if pd.Series([False, True, False]).any(): print("I am any")
Its output is as follows −
I am any
To evaluate single-element pandas objects in a Boolean context, use the method .bool() −
import pandas as pd print pd.Series([True]).bool()
Its output is as follows −
True
Bitwise Boolean
Bitwise Boolean operators like == and != will return a Boolean series, which is almost always what is required anyways.
import pandas as pd s = pd.Series(range(5)) print s==4
Its output is as follows −
0 False 1 False 2 False 3 False 4 True dtype: bool
isin Operation
This returns a Boolean series showing whether each element in the Series is exactly contained in the passed sequence of values.
import pandas as pd s = pd.Series(list('abc')) s = s.isin(['a', 'c', 'e']) print s
Its output is as follows −
0 True 1 False 2 True dtype: bool
Reindexing vs ix Gotcha
Many users will find themselves using the ix indexing capabilities as a concise means of selecting data from a Pandas object −
import pandas as pd import numpy as np df = pd.DataFrame(np.random.randn(6, 4), columns=['one', 'two', 'three', 'four'],index=list('abcdef')) print df print df.ix[['b', 'c', 'e']]
Its output is as follows −
one two three four a -1.582025 1.335773 0.961417 -1.272084 b 1.461512 0.111372 -0.072225 0.553058 c -1.240671 0.762185 1.511936 -0.630920 d -2.380648 -0.029981 0.196489 0.531714 e 1.846746 0.148149 0.275398 -0.244559 f -1.842662 -0.933195 2.303949 0.677641 one two three four b 1.461512 0.111372 -0.072225 0.553058 c -1.240671 0.762185 1.511936 -0.630920 e 1.846746 0.148149 0.275398 -0.244559
This is, of course, completely equivalent in this case to using the reindex method −
import pandas as pd import numpy as np df = pd.DataFrame(np.random.randn(6, 4), columns=['one', 'two', 'three', 'four'],index=list('abcdef')) print df print df.reindex(['b', 'c', 'e'])
Its output is as follows −
one two three four a 1.639081 1.369838 0.261287 -1.662003 b -0.173359 0.242447 -0.494384 0.346882 c -0.106411 0.623568 0.282401 -0.916361 d -1.078791 -0.612607 -0.897289 -1.146893 e 0.465215 1.552873 -1.841959 0.329404 f 0.966022 -0.190077 1.324247 0.678064 one two three four b -0.173359 0.242447 -0.494384 0.346882 c -0.106411 0.623568 0.282401 -0.916361 e 0.465215 1.552873 -1.841959 0.329404
Some might conclude that ix and reindex are 100% equivalent based on this. This is true except in the case of integer indexing. For example, the above operation can alternatively be expressed as −
import pandas as pd import numpy as np df = pd.DataFrame(np.random.randn(6, 4), columns=['one', 'two', 'three', 'four'],index=list('abcdef')) print df print df.ix[[1, 2, 4]] print df.reindex([1, 2, 4])
Its output is as follows −
one two three four a -1.015695 -0.553847 1.106235 -0.784460 b -0.527398 -0.518198 -0.710546 -0.512036 c -0.842803 -1.050374 0.787146 0.205147 d -1.238016 -0.749554 -0.547470 -0.029045 e -0.056788 1.063999 -0.767220 0.212476 f 1.139714 0.036159 0.201912 0.710119 one two three four b -0.527398 -0.518198 -0.710546 -0.512036 c -0.842803 -1.050374 0.787146 0.205147 e -0.056788 1.063999 -0.767220 0.212476 one two three four 1 NaN NaN NaN NaN 2 NaN NaN NaN NaN 4 NaN NaN NaN NaN
It is important to remember that reindex is strict label indexing only. This can lead to some potentially surprising results in pathological cases where an index contains, say, both integers and strings.