I have table with last 20 columns with numbers with % sign, how can I format the negative numbers to have red background and positive to have green. My table is a filtered table and has the following code
return dash_table.DataTable(
id='result-table',
columns=[{'name': col, 'id': col, 'type': 'numeric', 'format': Format(precision=2, scheme=Scheme.fixed, symbol=Symbol.yes, symbol_suffix='%')}
if col in filtered_df.columns[-20:] else {'name': col, 'id': col}
for col in filtered_df.columns],
data=filtered_df.to_dict('records'),
style_header=style_header,
style_cell=style_data
)
I tried the following but all of the cells in last 20 columns are green
style_data_conditional = [{'if': {'column_id': col}, 'backgroundColor': 'green', 'color': 'white'}
if col in filtered_df.columns[-20:] and col != 'Date' and row[col] >= 0 else
{'if': {'column_id': col}, 'backgroundColor': 'red', 'color': 'white'}
if col in filtered_df.columns[-20:] and col != 'Date' and row[col] < 0 else {}
for col in filtered_df.columns for row in filtered_df.to_dict('records')]
return dash_table.DataTable(
id='result-table',
columns=[{'name': col, 'id': col, 'type': 'numeric', 'format': Format(precision=2, scheme=Scheme.fixed, symbol=Symbol.yes, symbol_suffix='%')}
if col in filtered_df.columns[-20:] else {'name': col, 'id': col}
for col in filtered_df.columns],
data=filtered_df.to_dict('records'),
style_header=style_header,
style_cell=style_data,
style_data_conditional=style_data_conditional
)
Related
Given the following dataframe
col1 col2
1 ('A->B', 'B->C')
2 ('A->D', 'D->C', 'C->F')
3 ('A->K', 'K->M', 'M->P')
...
I want to convert this to the following format
col1 col2
1 'A-B-C'
2 'A-D-C-F'
3 'A-K-M-P'
...
Each sequence shows an arc within a path. Hence, the sequence is like (a,b), (b,c), (c,d) ...
def merge_values(val):
val = [x.split('->') for x in val]
out = []
for char in val:
out.append(char[0])
out.append(val[-1][1])
return '-'.join(out)
df['col2'] = df['col2'].apply(merge_values)
print(df)
Output:
col1 col2
0 1 A-B-C
1 2 A-D-C-F
2 3 A-K-M-P
Given
df = pd.DataFrame({
'col1': [1, 2, 3],
'col2': [
('A->B', 'B->C'),
('A->D', 'D->C', 'C->F'),
('A->K', 'K->M', 'M->P'),
],
})
You can do:
def combine(t, old_sep='->', new_sep='-'):
if not t: return ''
if type(t) == str: t = [t]
tokens = [x.partition(old_sep)[0] for x in t]
tokens += t[-1].partition(old_sep)[-1]
return new_sep.join(tokens)
df['col2'] = df['col2'].apply(combine)
I want to add relationships to column 'relations' based on rel_list. Specifically, for each tuple, i.e. ('a', 'b'), I want to replace the relationships column value '' with 'b' in the first row, but no duplicate, meaning that for the 2nd row, don't replace '' with 'a', since they are considered as duplicated. The following code doesn't work fully correct:
import pandas as pd
data = {
"names": ['a', 'b', 'c', 'd'],
"ages": [50, 40, 45, 20],
"relations": ['', '', '', '']
}
rel_list = [('a', 'b'), ('a', 'c'), ('c', 'd')]
df = pd.DataFrame(data)
for rel_tuple in rel_list:
head = rel_tuple[0]
tail = rel_tuple[1]
df.loc[df.names == head, 'relations'] = tail
print(df)
The current result of df is:
names ages relations
0 a 50 c
1 b 40
2 c 45 d
3 d 20
However, the correct one is:
names ages relations
0 a 50 b
0 a 50 c
1 b 40
2 c 45 d
3 d 20
There are new rows that need to be added. The 2nd row in this case, like above. How to do that?
You can craft a dataframe and merge:
(df.drop('relations', axis=1)
.merge(pd.DataFrame(rel_list, columns=['names', 'relations']),
on='names',
how='outer'
)
# .fillna('') # uncomment to replace NaN with empty string
)
Output:
names ages relations
0 a 50 b
1 a 50 c
2 b 40 NaN
3 c 45 d
4 d 20 NaN
Instead of updating df you can create a new one and add relations row by row:
import pandas as pd
data = {
"names": ['a', 'b', 'c', 'd'],
"ages": [50, 40, 45, 20],
"relations": ['', '', '', '']
}
rel_list = [('a', 'b'), ('a', 'c'), ('c', 'd')]
df = pd.DataFrame(data)
new_df = pd.DataFrame(data)
new_df.loc[:, 'relations'] = ''
for head, tail in rel_list:
new_row = df[df.names == head]
new_row.loc[:,'relations'] = tail
new_df = new_df.append(new_row)
print(new_df)
Output:
names ages relations
0 a 50
1 b 40
2 c 45
3 d 20
0 a 50 b
0 a 50 c
2 c 45 d
Then, if needed, in the end you can delete all rows without value in 'relations':
new_df = new_df[new_df['relations']!='']
Have any way to use
df = pd.read_excel(r'a.xlsx')
df2 = df.groupby(by=["col"], as_index=False).mean()
Include new column with number of rows grouped in each row?
in absence of sample data, I'm assuming you have multiple numeric columns
can use apply() to then calculate all means and append len() to this series
df = pd.DataFrame(
{
"col": np.random.choice(list("ABCD"), 200),
"val": np.random.uniform(1, 5, 200),
"val2": np.random.uniform(5, 10, 200),
}
)
df2 = df.groupby(by=["col"], as_index=False).apply(
lambda d: d.select_dtypes("number").mean().append(pd.Series({"len": len(d)}))
)
df2
col
val
val2
len
0
A
3.13064
7.63837
42
1
B
3.1057
7.50656
44
2
C
3.0111
7.82628
54
3
D
3.20709
7.32217
60
comment code
def w_avg(df, values, weights, exp):
d = df[values]
w = df[weights] ** exp
return (d * w).sum() / w.sum()
dfg1 = pd.DataFrame(
{
"Jogador": np.random.choice(list("ABCD"), 200),
"Evento": np.random.choice(list("XYZ"),200),
"Rating Calculado BW": np.random.uniform(1, 5, 200),
"Lances": np.random.uniform(5, 10, 200),
}
)
dfg = dfg1.groupby(by=["Jogador", "Evento"]).apply(
lambda dfg1: dfg1.select_dtypes("number")
.agg(lambda d: w_avg(dfg1, "Rating Calculado BW", "Lances", 1))
.append(pd.Series({"len": len(dfg1)}))
)
dfg
df = pd.DataFrame({'ID1' : ['A' , 'A', 'B'],
'ID2' : ['C' , 'D', 'E'],
'bool' : [True, True, False]})
df_agg = df.groupby('ID1').agg(lambda x: ';'.join(set(x))).reset_index()
bool_col = df.drop_duplicates(subset=['ID1'])[['bool']].reset_index(drop=True)
final_df = pd.concat([df_agg, bool_col], axis=1)
I want to string concat ID2 when ID1 is duplicated, bot only want to keep the largest value (True) for col bool. I almost have it here, but there has to be a better way
You can pass agg with dict
out = df.groupby('ID1',as_index=False).agg({'ID2': lambda x : ','.join(set(x)),'bool' : 'last'})
Out[322]:
ID1 ID2 bool
0 A C,D True
1 B E False
You're searching for the largest value of bool column; so i'll go with this approach:
df1 = df.groupby('ID1').agg({
'ID2': lambda x: ','.join(set(x)),
'bool': 'max'
}).reset_index()
print(df1)
Output:
ID1 ID2 bool
0 A C,D True
1 B E False
I have a DataFrame with a MultiIndex with 3 levels:
id foo bar col1
0 1 a -0.225873
2 a -0.275865
2 b -1.324766
3 1 a -0.607122
2 a -1.465992
2 b -1.582276
3 b -0.718533
7 1 a -1.904252
2 a 0.588496
2 b -1.057599
3 a 0.388754
3 b -0.940285
Preserving the id index level, I want to sum along the foo and bar levels, but with different values for each id.
For example, for id = 0 I want to sum over foo = [1] and bar = [["a", "b"]], for id = 3 I want to sum over foo = [2] and bar = [["a", "b"]], and for id = 7 I want to sum over foo = [[1,2]] and bar = [["a"]]. Giving the result:
id col1
0 -0.225873
3 -3.048268
7 -1.315756
I have been trying something along these lines:
df.loc(axis = 0)[[(0, 1, ["a","b"]), (3, 2, ["a","b"]), (7, [1,2], "a")].sum()
Not sure if this is even possible. Any elegant solution (possibly removing the MultiIndex?) would be much appreciated!
The list of tuples is not the problem. The fact that each tuple does not correspond to a single index is the problem (Since a list isn't a valid key). If you want to index a Dataframe like this, you need to expand the lists inside each tuple to their own entries.
Define your options like the following list of dictionaries, then transform using a list comprehension and index using all individual entries.
d = [
{
'id': 0,
'foo': [1],
'bar': ['a', 'b']
},
{
'id': 3,
'foo': [2],
'bar': ['a', 'b']
},
{
'id': 7,
'foo': [1, 2],
'bar': ['a']
},
]
all_idx = [
(el['id'], i, j)
for el in d
for i in el['foo']
for j in el['bar']
]
# [(0, 1, 'a'), (0, 1, 'b'), (3, 2, 'a'), (3, 2, 'b'), (7, 1, 'a'), (7, 2, 'a')]
df.loc[all_idx].groupby(level=0).sum()
col1
id
0 -0.225873
3 -3.048268
7 -1.315756
A more succinct solution using slicers:
sections = [(0, 1, slice(None)), (3, 2, slice(None)), (7, slice(1,2), "a")]
pd.concat(df.loc[s] for s in sections).groupby("id").sum()
col1
id
0 -0.225873
3 -3.048268
7 -1.315756
Two things to note:
This may be less memory-efficient than the accepted answer since pd.concat creates a new DataFrame.
The slice(None)'s are mandatory, otherwise the index columns of the df.loc[s]'s mismatch when calling pd.concat.