更改绘图中间的线条颜色 - plotly
change line color in middle of plot - plotly
从包含具有 13 个值的行的 csv 中创建 12 个子图。最后一个值表示该行中的数据是从这一点开始估计的，直到状态在后面的行中变回。
尝试制作绘制一条线的折线图，当状态变为 1 时变为红色，然后当状态变为 0 时返回原始颜色。这可能吗？
with lock:
df = pd.read_csv(OCcsvFile, delimiter=',')
# plotly setup
plot_rows = 4
plot_cols = 3
# Create plot figure
fig = make_subplots(rows=plot_rows, cols=plot_cols, subplot_titles=("Header1", "Header2", "Header3", "Header4", "Header5",
"Header6", "Header7", "Header8", "Header9", "Header10",
"Header11", "Header12"))
# add traces
x = 1 # column counter
for i in range(1, plot_rows+1):
for j in range(1, plot_cols+1):
#print(str(i)+ ', ' + str(j))
fig.add_trace(go.Scatter(x=df.iloc[:, 0], y=df.iloc[:, x],
name=df.columns[x],
mode='lines'),
row=i,
col=j)
x = x+1
- 采用重塑数据框的方法为 plotly express
做好准备
- 已经根据描述计算出起始数据框，有样本会更好
import pandas as pd
import plotly.express as px
df = pd.DataFrame(
np.random.randint(1, 10, [100, 12]), columns=[f"c{i+1}" for i in range(12)]
).assign(status=np.repeat(np.random.randint(0, 2, 20), 5))
# restructure dataframe for px
# 1. preserve status in index
# 2. make columns another level of index
# 3. make index columns and make column names meaningful
dfp = (
df.set_index("status", append=True)
.stack()
.reset_index()
.rename(columns={"level_0": "x", "level_2": "facet", 0: "value"})
)
# make sure missing values are present as NaN
dfp = dfp.merge(
pd.DataFrame(index=pd.MultiIndex.from_product(
[dfp["x"].unique(), dfp["facet"].unique(), dfp["status"].unique()]
)),
left_on=["x", "facet", "status"],
right_index=True,
how="right"
)
# now it's a very simple plot
px.line(dfp, x="x", y="value", color="status", facet_col="facet", facet_col_wrap=4)
预期结构或 df
- 13 列，最后一列表示状态
c1
c2
c3
c4
c5
c6
c7
c8
c9
c10
c11
c12
status
10
7
3
1
7
2
8
1
3
9
6
3
8
1
11
4
5
8
9
5
4
3
6
3
7
4
8
1
12
6
3
2
6
5
6
4
3
5
3
9
7
1
13
4
2
4
8
6
3
3
5
8
8
1
4
1
14
4
9
9
3
1
8
2
5
1
5
1
4
1
15
4
9
6
2
9
4
1
6
6
1
6
1
0
16
8
5
9
7
7
3
1
1
2
5
2
9
0
17
6
1
4
2
8
5
9
8
2
4
8
4
0
18
1
6
1
3
8
5
5
9
8
9
2
9
0
19
1
4
1
1
7
8
2
3
5
6
6
4
0
从包含具有 13 个值的行的 csv 中创建 12 个子图。最后一个值表示该行中的数据是从这一点开始估计的，直到状态在后面的行中变回。
尝试制作绘制一条线的折线图，当状态变为 1 时变为红色，然后当状态变为 0 时返回原始颜色。这可能吗？
with lock:
df = pd.read_csv(OCcsvFile, delimiter=',')
# plotly setup
plot_rows = 4
plot_cols = 3
# Create plot figure
fig = make_subplots(rows=plot_rows, cols=plot_cols, subplot_titles=("Header1", "Header2", "Header3", "Header4", "Header5",
"Header6", "Header7", "Header8", "Header9", "Header10",
"Header11", "Header12"))
# add traces
x = 1 # column counter
for i in range(1, plot_rows+1):
for j in range(1, plot_cols+1):
#print(str(i)+ ', ' + str(j))
fig.add_trace(go.Scatter(x=df.iloc[:, 0], y=df.iloc[:, x],
name=df.columns[x],
mode='lines'),
row=i,
col=j)
x = x+1
- 采用重塑数据框的方法为 plotly express 做好准备
- 已经根据描述计算出起始数据框，有样本会更好
import pandas as pd
import plotly.express as px
df = pd.DataFrame(
np.random.randint(1, 10, [100, 12]), columns=[f"c{i+1}" for i in range(12)]
).assign(status=np.repeat(np.random.randint(0, 2, 20), 5))
# restructure dataframe for px
# 1. preserve status in index
# 2. make columns another level of index
# 3. make index columns and make column names meaningful
dfp = (
df.set_index("status", append=True)
.stack()
.reset_index()
.rename(columns={"level_0": "x", "level_2": "facet", 0: "value"})
)
# make sure missing values are present as NaN
dfp = dfp.merge(
pd.DataFrame(index=pd.MultiIndex.from_product(
[dfp["x"].unique(), dfp["facet"].unique(), dfp["status"].unique()]
)),
left_on=["x", "facet", "status"],
right_index=True,
how="right"
)
# now it's a very simple plot
px.line(dfp, x="x", y="value", color="status", facet_col="facet", facet_col_wrap=4)
预期结构或 df
- 13 列，最后一列表示状态
| c1 | c2 | c3 | c4 | c5 | c6 | c7 | c8 | c9 | c10 | c11 | c12 | status | |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| 10 | 7 | 3 | 1 | 7 | 2 | 8 | 1 | 3 | 9 | 6 | 3 | 8 | 1 |
| 11 | 4 | 5 | 8 | 9 | 5 | 4 | 3 | 6 | 3 | 7 | 4 | 8 | 1 |
| 12 | 6 | 3 | 2 | 6 | 5 | 6 | 4 | 3 | 5 | 3 | 9 | 7 | 1 |
| 13 | 4 | 2 | 4 | 8 | 6 | 3 | 3 | 5 | 8 | 8 | 1 | 4 | 1 |
| 14 | 4 | 9 | 9 | 3 | 1 | 8 | 2 | 5 | 1 | 5 | 1 | 4 | 1 |
| 15 | 4 | 9 | 6 | 2 | 9 | 4 | 1 | 6 | 6 | 1 | 6 | 1 | 0 |
| 16 | 8 | 5 | 9 | 7 | 7 | 3 | 1 | 1 | 2 | 5 | 2 | 9 | 0 |
| 17 | 6 | 1 | 4 | 2 | 8 | 5 | 9 | 8 | 2 | 4 | 8 | 4 | 0 |
| 18 | 1 | 6 | 1 | 3 | 8 | 5 | 5 | 9 | 8 | 9 | 2 | 9 | 0 |
| 19 | 1 | 4 | 1 | 1 | 7 | 8 | 2 | 3 | 5 | 6 | 6 | 4 | 0 |