bbd*_*108 10 python plotly plotly-python
样本数据如下:
unique_list = ['home0', 'page_a0', 'page_b0', 'page_a1', 'page_b1',
'page_c1', 'page_b2', 'page_a2', 'page_c2', 'page_c3']
sources = [0, 0, 1, 2, 2, 3, 3, 4, 4, 7, 6]
targets = [3, 4, 4, 3, 5, 6, 8, 7, 8, 9, 9]
values = [2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2]
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使用文档中的示例代码
fig = go.Figure(data=[go.Sankey(
node = dict(
pad = 15,
thickness = 20,
line = dict(color = "black", width = 0.5),
label = unique_list,
color = "blue"
),
link = dict(
source = sources,
target = targets,
value = values
))])
fig.show()
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这将输出以下桑基图
但是,我想获取同一垂直列中以相同数字结尾的所有值,就像最左边的列的所有节点都以 0 结尾一样。我在文档中看到可以移动节点位置,但是我想知道除了手动输入 x 和 y 值之外是否有更干净的方法来做到这一点。任何帮助表示赞赏。
ves*_*and 15
在和中go.Sankey()设置arrangement='snap'并调整 x 和 y 位置。以下设置将根据要求放置您的节点。x=<list>y=<list>
阴谋:
请注意,本示例中未明确设置 y 值。一旦有多个节点具有相同的 x 值,y 值将自动调整,以使所有节点显示在同一垂直位置。如果您确实想明确设置所有位置,只需设置arrangement='fixed'
编辑:
我添加了一个自定义函数nodify(),该函数将相同的 x 位置分配给具有共同结尾(例如'0'中)的标签名称['home0', 'page_a0', 'page_b0']。page_c1现在,如果您更改为示例,page_c2您将得到以下结果:
完整代码:
import plotly.graph_objects as go
unique_list = ['home0', 'page_a0', 'page_b0', 'page_a1', 'page_b1',
'page_c1', 'page_b2', 'page_a2', 'page_c2', 'page_c3']
sources = [0, 0, 1, 2, 2, 3, 3, 4, 4, 7, 6]
targets = [3, 4, 4, 3, 5, 6, 8, 7, 8, 9, 9]
values = [2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2]
def nodify(node_names):
node_names = unique_list
# uniqe name endings
ends = sorted(list(set([e[-1] for e in node_names])))
# intervals
steps = 1/len(ends)
# x-values for each unique name ending
# for input as node position
nodes_x = {}
xVal = 0
for e in ends:
nodes_x[str(e)] = xVal
xVal += steps
# x and y values in list form
x_values = [nodes_x[n[-1]] for n in node_names]
y_values = [0.1]*len(x_values)
return x_values, y_values
nodified = nodify(node_names=unique_list)
# plotly setup
fig = go.Figure(data=[go.Sankey(
arrangement='snap',
node = dict(
pad = 15,
thickness = 20,
line = dict(color = "black", width = 0.5),
label = unique_list,
color = "blue",
x=nodified[0],
y=nodified[1]
),
link = dict(
source = sources,
target = targets,
value = values
))])
fig.show()
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