I have a dataframe:
ID
Name
1
A
2
B
3
C
I defined a list:
mylist =[A,C]
If I want to extract only the rows where Name is equal to A and C (namely, mylist), I am trying to use the following code:
df_new = df[(df['Name'].isin(mylist))]
>>> df_new
As result, I get an empty table.
Any suggestion regarding why I get this error?
Just remove the additional open bracket before the df['Name']
df_new = df[df['Name'].isin(lst)]
Found the solution, It was a problem related to the list that caused the result of the empty table.
The format of the list should be:
mylist =['A','C']
instead of
mylist =[A,C]
You could use .loc and lambda as it’s more readable
import pandas as pd
dataf = pd.DataFrame({'ID':[1,2,3],'Name':['A','B','C']})
names = ['A','C']
# lock rows where column Name in names
df = dataf.loc[lambda d: d['Name'].isin(names)]
print(df)
Related
I'm new to python and especially to pandas so I don't really know what I'm doing. I have 10 columns with 100000 rows and 4 letter strings. I need to filter out rows which don't contain 'DDD' in all of the columns/rows.
I tried to do it with iloc and loc, but it doesn't work:
import pandas as pd
df = pd.read_csv("data_3.csv", delimiter = '!')
df.iloc[:,10:20].str.contains('DDD', regex= False, na = False)
df.head()
It returns me an error: 'DataFrame' object has no attribute 'str'
I suggest doing it without a for loop like this:
df[df.apply(lambda x: x.str.contains('DDD')).all(axis=1)]
To select only string columns
df[df.select_dtypes(include='object').apply(lambda x: x.str.contains('DDD')).all(axis=1)]
To select only some string columns
selected_cols = ['A','B']
df[df[selected_cols].apply(lambda x: x.str.contains('DDD')).all(axis=1)]
You can do this but if your all column type is StringType:
for column in foo.columns:
df = df[~df[c].str.contains('DDD')]
You can use str.contains, but only on Series not on DataFrames. So to use it we look at each column (which is a series) one by one by for looping over them:
>>> import pandas as pd
>>> df = pd.DataFrame([['DDDA', 'DDDB', 'DDDC', 'DDDD'],
['DDDE', 'DDDF', 'DDDG', 'DHDD'],
['DDDI', 'DDDJ', 'DDDK', 'DDDL'],
['DMDD', 'DNDN', 'DDOD', 'DDDP']],
columns=['A', 'B', 'C', 'D'])
>>> for column in df.columns:
df = df[df[column].str.contains('DDD')]
In our for loop we're overwriting the DataFrame df with df where the column contains 'DDD'. By looping over each column we cut out rows that don't contain 'DDD' in that column until we've looked in all of our columns, leaving only rows that contain 'DDD' in every column.
This gives you:
>>> print(df)
A B C D
0 DDDA DDDB DDDC DDDD
2 DDDI DDDJ DDDK DDDL
As you're only looping over 10 columns this shouldn't be too slow.
Edit: You should probably do it without a for loop as explained by Christian Sloper as it's likely to be faster, but I'll leave this up as it's slightly easier to understand without knowledge of lambda functions.
I have a data frame in pyspark with more than 100 columns. What I want to do is for all the column names I would like to add back ticks(`) at the start of the column name and end of column name.
For example:
column name is testing user. I want `testing user`
Is there a method to do this in pyspark/python. when we apply the code it should return a data frame.
Use list comprehension in python.
from pyspark.sql import functions as F
df = ...
df_new = df.select([F.col(c).alias("`"+c+"`") for c in df.columns])
This method also gives you the option to add custom python logic within the alias() function like: "prefix_"+c+"_suffix" if c in list_of_cols_to_change else c
To add prefix or suffix:
Refer df.columns for list of columns ([col_1, col_2...]). This is the dataframe, for which we want to suffix/prefix column.
df.columns
Iterate through above list and create another list of columns with alias that can used inside select expression.
from pyspark.sql.functions import col
select_list = [col(col_name).alias("prefix_" + col_name) for col_name in df.columns]
When using inside select, do not forget to unpack list with asterisk(*). We can assign it back to same or different df for use.
df.select(*select_list).show()
df = df.select(*select_list)
df.columns will now return list of new columns(aliased).
If you would like to add a prefix or suffix to multiple columns in a pyspark dataframe, you could use a for loop and .withColumnRenamed().
As an example, you might like:
def add_prefix(sdf, prefix):
for c in sdf.columns:
sdf = sdf.withColumnRenamed(c, '{}{}'.format(prefix, c))
return sdf
You can amend sdf.columns as you see fit.
You can use withColumnRenamed method of dataframe in combination with na to create new dataframe
df.na.withColumnRenamed('testing user', '`testing user`')
edit : suppose you have list of columns, you can do like -
old = "First Last Age"
new = ["`"+field+"`" for field in old.split()]
df.rdd.toDF(new)
output :
DataFrame[`First`: string, `Last`: string, `Age`: string]
here is how one can solve the similar problems:
df.select([col(col_name).alias('prefix' + col_name + 'suffix') for col_name in df])
I had a dataframe that I duplicated twice then joined together. Since both had the same columns names I used :
df = reduce(lambda df, idx: df.withColumnRenamed(list(df.schema.names)[idx],
list(df.schema.names)[idx] + '_prec'),
range(len(list(df.schema.names))),
df)
Every columns in my dataframe then had the '_prec' suffix which allowed me to do sweet stuff
I have a table with two columns and i want to combine the text with the same id
import pandas as pd
df = DataFrame({'id':[101453,101465,101478,101453,101465,101465], 'text' :['this','is','a','test','string','one']})
I need a result like this:
df = DataFrame({'id':[101453,101465,101478], 'text':['this test','is string one','a']})
Use groupby with apply join:
print (df.groupby('id')['text'].apply(' '.join).reset_index())
id text
0 101453 this test
1 101465 is string one
2 101478 a
df['id'] = sorted(list(set(df['id'])))
set() removes all equal elements. Then return it to list(). And sort it if you need.
How to select all columns that have header names starting with "durations" or "shape"? (instead of defining a long list of column names). I need to select these columns and substitute blank fields by 0.
column_names = ['durations.blockMinutes_x',
'durations.scheduledBlockMinutes_y']
data[column_names] = data[column_names].fillna(0)
You could use str methods of dataframe startwith:
df = data[data.columns[data.columns.str.startwith('durations') | data.columns.str.startwith('so')]]
df.fillna(0)
Or you could use contains method:
df = data.iloc[:, data.columns.str.contains('durations.*'|'shape.*') ]
df.fillna(0)
I would use the select method:
df.select(lambda c: c.startwith('durations') or c.startswith('shape'), axis=1)
Use my_dataframe.columns.values.tolist() to get the column names (based on Get list from pandas DataFrame column headers):
column_names = [x for x in data.columns.values.tolist() if x.startswith("durations") or x.startswith("shape")]
A simple and easy way
data[data.filter(regex='durations|shape').columns].fillna(0)
Sample Screenshot
I have a dataframe (df) and want to print the unique values from each column in the dataframe.
I need to substitute the variable (i) [column name] into the print statement
column_list = df.columns.values.tolist()
for column_name in column_list:
print(df."[column_name]".unique()
Update
When I use this: I get "Unexpected EOF Parsing" with no extra details.
column_list = sorted_data.columns.values.tolist()
for column_name in column_list:
print(sorted_data[column_name].unique()
What is the difference between your syntax YS-L (above) and the below:
for column_name in sorted_data:
print(column_name)
s = sorted_data[column_name].unique()
for i in s:
print(str(i))
It can be written more concisely like this:
for col in df:
print(df[col].unique())
Generally, you can access a column of the DataFrame through indexing using the [] operator (e.g. df['col']), or through attribute (e.g. df.col).
Attribute accessing makes the code a bit more concise when the target column name is known beforehand, but has several caveats -- for example, it does not work when the column name is not a valid Python identifier (e.g. df.123), or clashes with the built-in DataFrame attribute (e.g. df.index). On the other hand, the [] notation should always work.
Most upvoted answer is a loop solution, hence adding a one line solution using pandas apply() method and lambda function.
print(df.apply(lambda col: col.unique()))
This will get the unique values in proper format:
pd.Series({col:df[col].unique() for col in df})
If you're trying to create multiple separate dataframes as mentioned in your comments, create a dictionary of dataframes:
df_dict = dict(zip([i for i in df.columns] , [pd.DataFrame(df[i].unique(), columns=[i]) for i in df.columns]))
Then you can access any dataframe easily using the name of the column:
df_dict[column name]
We can make this even more concise:
df.describe(include='all').loc['unique', :]
Pandas describe gives a few key statistics about each column, but we can just grab the 'unique' statistic and leave it at that.
Note that this will give a unique count of NaN for numeric columns - if you want to include those columns as well, you can do something like this:
df.astype('object').describe(include='all').loc['unique', :]
I was seeking for a solution to this problem as well, and the code below proved to be more helpful in my situation,
for col in df:
print(col)
print(df[col].unique())
print('\n')
It gives something like below:
Fuel_Type
['Diesel' 'Petrol' 'CNG']
HP
[ 90 192 69 110 97 71 116 98 86 72 107 73]
Met_Color
[1 0]
The code below could provide you a list of unique values for each field, I find it very useful when you want to take a deeper look at the data frame:
for col in list(df):
print(col)
print(df[col].unique())
You can also sort the unique values if you want them to be sorted:
import numpy as np
for col in list(df):
print(col)
print(np.sort(df[col].unique()))
cu = []
i = []
for cn in card.columns[:7]:
cu.append(card[cn].unique())
i.append(cn)
pd.DataFrame( cu, index=i).T
Simply do this:
for i in df.columns:
print(df[i].unique())
Or in short it can be written as:
for val in df['column_name'].unique():
print(val)
Even better. Here's code to view all the unique values as a dataframe column-wise transposed:
columns=[*df.columns]
unique_values={}
for i in columns:
unique_values[i]=df[i].unique()
unique=pd.DataFrame(dict([ (k,pd.Series(v)) for k,v in unique_vals.items() ]))
unique.fillna('').T
This solution constructs a dataframe of unique values with some stats and gracefully handles any unhashable column types.
Resulting dataframe columns are: col, unique_len, df_len, perc_unique, unique_values
df_len = len(df)
unique_cols_list = []
for col in df:
try:
unique_values = df[col].unique()
unique_len = len(unique_values)
except TypeError: # not all cols are hashable
unique_values = ""
unique_len = -1
perc_unique = unique_len*100/df_len
unique_cols_list.append((col, unique_len, df_len, perc_unique, unique_values))
df_unique_cols = pd.DataFrame(unique_cols_list, columns=["col", "unique_len", "df_len", "perc_unique", "unique_values"])
df_unique_cols = df_unique_cols[df_unique_cols["unique_len"] > 0].sort_values("unique_len", ascending=False)
print(df_unique_cols)
The best way to do that:
Series.unique()
For example students.age.unique() the output will be the different values that occurred in the age column of the students' data frame.
To get only the number of how many different values:
Series.nunique()