Is it possible to append to an empty data frame that doesn't contain any indices or columns?
I have tried to do this, but keep getting an empty dataframe at the end.
e.g.
import pandas as pd
df = pd.DataFrame()
data = ['some kind of data here' --> I have checked the type already, and it is a dataframe]
df.append(data)
The result looks like this:
Empty DataFrame
Columns: []
Index: []
This should work:
>>> df = pd.DataFrame()
>>> data = pd.DataFrame({"A": range(3)})
>>> df = df.append(data)
>>> df
A
0 0
1 1
2 2
Since the append doesn't happen in-place, so you'll have to store the output if you want it:
>>> df = pd.DataFrame()
>>> data = pd.DataFrame({"A": range(3)})
>>> df.append(data) # without storing
>>> df
Empty DataFrame
Columns: []
Index: []
>>> df = df.append(data)
>>> df
A
0 0
1 1
2 2
And if you want to add a row, you can use a dictionary:
df = pd.DataFrame()
df = df.append({'name': 'Zed', 'age': 9, 'height': 2}, ignore_index=True)
which gives you:
age height name
0 9 2 Zed
You can concat the data in this way:
InfoDF = pd.DataFrame()
tempDF = pd.DataFrame(rows,columns=['id','min_date'])
InfoDF = pd.concat([InfoDF,tempDF])
The answers are very useful, but since pandas.DataFrame.append was deprecated (as already mentioned by various users), and the answers using pandas.concat are not "Runnable Code Snippets" I would like to add the following snippet:
import pandas as pd
df = pd.DataFrame(columns =['name','age'])
row_to_append = pd.DataFrame([{'name':"Alice", 'age':"25"},{'name':"Bob", 'age':"32"}])
df = pd.concat([df,row_to_append])
So df is now:
name age
0 Alice 25
1 Bob 32
pandas.DataFrame.append Deprecated since version 1.4.0: Use concat() instead.
Therefore:
df = pd.DataFrame() # empty dataframe
df2 = pd..DataFrame(...) # some dataframe with data
df = pd.concat([df, df2])
Related
When I run the following code:
import pandas as pd
data = {'age':{'sam': 1,
'rye': 3,
'lori':8,
'chris':11,
'sara':3}}
df = pd.DataFrame()
df = pd.concat([df, data])
I get this error:
TypeError: cannot concatenate object of type '<class 'dict'>'; only Series and DataFrame objs are valid
I'm wondering if this action is possible or if I'm doing something incorrectly.
You just need to make data into a DataFrame first:
>>> df = pd.concat([df, pd.DataFrame(data)])
>>> df
age
chris 11
lori 8
rye 3
sam 1
sara 3
You could then concat further dictionaries to df in a loop.
I want to replace a row in a csv file with a variable. The row itself also has to be a variable. The following code is an example:
import pandas as pd
# sample dataframe
df = pd.DataFrame({'A': ['a','b','c'], 'B':['b','c','d']})
print("Original DataFrame:\n", df)
x = 1
y = 12698
df_rep = df.replace([int(x),1], y)
print("\nAfter replacing:\n", df_rep)
This can be done using pandas indexing eg df.iloc[row_num, col_num].
#update df
df.iloc[x,1]=y
#print df
print(df)
A B
0 a b
1 b 12698
2 c d
I have two large DataFrames that I don't want to make copies of, but want to apply the same change to. How can I do this properly? For example, this is similar to what I want to do, but on a smaller scale. This only creates the temporary variable df that gives the result of each DataFrame, but I want both DataFrames to be themselves changed:
import pandas as pd
df1 = pd.DataFrame({'a':[1,2,3]})
df2 = pd.DataFrame({'a':[0,1,5,7]})
for df in [df1, df2]:
df = df[df['a'] < 3]
We can do query with inplace
df1 = pd.DataFrame({'a':[1,2,3]})
df2 = pd.DataFrame({'a':[0,1,5,7]})
for df in [df1, df2]:
df.query('a<3',inplace=True)
df1
a
0 1
1 2
df2
a
0 0
1 1
Don't think this is the best solution, but should do the job.
import pandas as pd
df1 = pd.DataFrame({'a':[1,2,3]})
df2 = pd.DataFrame({'a':[0,1,5,7]})
dfs = [df1, df2]
for i, df in enumerate(dfs):
dfs[i] = df[df['a'] < 3]
dfs[0]
a
0 1
1 2
I have a dataframe with CSVs in language column
Name Language
0 A French,Espanol
1 B Deutsch,English
I wish to transform the above dataframe as below
Name Language
0 A French
1 A Espanol
2 B Deutsch
3 B English
I tried the below code but couldn't accomplish
df=df.join(df.pop('Language').str.extractall(',$')[0] .reset_index(level=1,drop=True) .rename('Language')) .reset_index(drop=True)
pandas.DataFrame.explode should be suited for that task. Combine it with pandas.DataFrame.assign to get the desired column:
import pandas as pd
df = pd.DataFrame({'Name':['A', 'B'], 'Language': ['French,Espanol', 'Deutsch,English']})
df = df.assign(Language=df['Language'].str.split(',')).explode('Language')
# Name Language
# 0 A French
# 0 A Espanol
# 1 B Deutsch
# 1 B English
First create a new dataframe with the same columns, then split second values and appent rows to the dataframe.
import pandas as pd
csv_df = pd.DataFrame([['1', '2,3'], ['2', '4,5']], columns=['Name', 'Language'])
df = pd.DataFrame(columns=['Name ', 'Language'])
for index, row in csv_df .iterrows():
name = row['Name']
s = row['Language']
txt = s.split(',')
for x in txt:
df = df.append(pd.Series([name, x], index=df.columns), ignore_index=True)
print(df)
I have a object of which type is Panda and the print(object) is giving below output
print(type(recomen_total))
print(recomen_total)
Output is
<class 'pandas.core.frame.Pandas'>
Pandas(Index=12, instrument_1='XXXXXX', instrument_2='XXXX', trade_strategy='XXX', earliest_timestamp='2016-08-02T10:00:00+0530', latest_timestamp='2016-08-02T10:00:00+0530', xy_signal_count=1)
I want to convert this obejct in pd.DataFrame, how i can do it ?
i tried pd.DataFrame(object), from_dict also , they are throwing error
Interestingly, it will not convert to a dataframe directly but to a series. Once this is converted to a series use the to_frame method of series to convert it to a DataFrame
import pandas as pd
df = pd.DataFrame({'col1': [1, 2], 'col2': [0.1, 0.2]},
index=['a', 'b'])
for row in df.itertuples():
print(pd.Series(row).to_frame())
Hope this helps!!
EDIT
In case you want to save the column names use the _asdict() method like this:
import pandas as pd
df = pd.DataFrame({'col1': [1, 2], 'col2': [0.1, 0.2]},
index=['a', 'b'])
for row in df.itertuples():
d = dict(row._asdict())
print(pd.Series(d).to_frame())
Output:
0
Index a
col1 1
col2 0.1
0
Index b
col1 2
col2 0.2
To create new DataFrame from itertuples namedtuple you can use list() or Series too:
import pandas as pd
# source DataFrame
df = pd.DataFrame({'a': [1,2], 'b':[3,4]})
# empty DataFrame
df_new_fromAppend = pd.DataFrame(columns=['x','y'], data=None)
for r in df.itertuples():
# create new DataFrame from itertuples() via list() ([1:] for skipping the index):
df_new_fromList = pd.DataFrame([list(r)[1:]], columns=['c','d'])
# or create new DataFrame from itertuples() via Series (drop(0) to remove index, T to transpose column to row)
df_new_fromSeries = pd.DataFrame(pd.Series(r).drop(0)).T
# or use append() to insert row into existing DataFrame ([1:] for skipping the index):
df_new_fromAppend.loc[df_new_fromAppend.shape[0]] = list(r)[1:]
print('df_new_fromList:')
print(df_new_fromList, '\n')
print('df_new_fromSeries:')
print(df_new_fromSeries, '\n')
print('df_new_fromAppend:')
print(df_new_fromAppend, '\n')
Output:
df_new_fromList:
c d
0 2 4
df_new_fromSeries:
1 2
0 2 4
df_new_fromAppend:
x y
0 1 3
1 2 4
To omit index, use param index=False (but I mostly need index for the iteration)
for r in df.itertuples(index=False):
# the [1:] needn't be used, for example:
df_new_fromAppend.loc[df_new_fromAppend.shape[0]] = list(r)
The following works for me:
import pandas as pd
df = pd.DataFrame({'col1': [1, 2], 'col2': [0.1, 0.2]}, index=['a', 'b'])
for row in df.itertuples():
row_as_df = pd.DataFrame.from_records([row], columns=row._fields)
print(row_as_df)
The result is:
Index col1 col2
0 a 1 0.1
Index col1 col2
0 b 2 0.2
Sadly, AFAIU, there's no simple way to keep column names, without explicitly utilizing "protected attributes" such as _fields.
With some tweaks in #Igor's answer
I concluded with this satisfactory code which preserved column names and used as less of pandas code as possible.
import pandas as pd
df = pd.DataFrame({'col1': [1, 2], 'col2': [0.1, 0.2]})
# Or initialize another dataframe above
# Get list of column names
column_names = df.columns.values.tolist()
filtered_rows = []
for row in df.itertuples(index=False):
# Some code logic to filter rows
filtered_rows.append(row)
# Convert pandas.core.frame.Pandas to pandas.core.frame.Dataframe
# Combine filtered rows into a single dataframe
concatinated_df = pd.DataFrame.from_records(filtered_rows, columns=column_names)
concatinated_df.to_csv("path_to_csv", index=False)
The result is a csv containing:
col1 col2
1 0.1
2 0.2
To convert a list of objects returned by Pandas .itertuples to a DataFrame, while preserving the column names:
# Example source DF
data = [['cheetah', 120], ['human', 44.72], ['dragonfly', 54]]
source_df = pd.DataFrame(data, columns=['animal', 'top_speed'])
animal top_speed
0 cheetah 120.00
1 human 44.72
2 dragonfly 54.00
Since Pandas does not recommended building DataFrames by adding single rows in a for loop, we will iterate and build the DataFrame at the end:
WOW_THAT_IS_FAST = 50
list_ = list()
for animal in source_df.itertuples(index=False, name='animal'):
if animal.top_speed > 50:
list_.append(animal)
Now build the DF in a single command and without manually recreating the column names.
filtered_df = pd.DataFrame(list_)
animal top_speed
0 cheetah 120.00
2 dragonfly 54.00