标签:matplotlib python
所以对于我一生,我无法弄清楚如何让标签显示小数位,而不仅仅是0,1,2
我需要它们是十进制形式,下面是我在python 3中的代码
#ROE and Growth
Tax_Burden = stock.loc['Net Income']/stock.loc['Pre-Tax Income']
Interest_Burden= stock.loc['Pre-Tax Income']/stock.loc['Operating Income']
Operating_Margin= stock.loc['Operating Income']/stock.loc['Revenue']
Asset_Turnover= stock.loc['Revenue']/stock.loc['Total Assets Average']
Leverage_Ratio= stock.loc['Total Assets Average']/stock.loc['Total Equity Average']
roe=Tax_Burden*Interest_Burden*Operating_Margin*Asset_Turnover*Leverage_Ratio
Growth = roe * (1-stock.loc['Dividend Payout Ratio'])
astart = 21
aend = 31
annual = [Operating_Margin[astart:aend],Tax_Burden[astart:aend],Interest_Burden[astart:aend],Asset_Turnover[astart:aend],Leverage_Ratio[astart:aend],roe[astart:aend],Growth[astart:aend]]
N = len(annual[0])
ind = np.arange(N) # the x locations for the groups
width = .12 # the width of the bars
fig, ax = plt.subplots(figsize=(20,10))
rects1 = ax.bar(ind, annual[0], width, color='y')
rects2 = ax.bar(ind+width, annual[1], width, color='r')
rects3 = ax.bar(ind+width*2, annual[2], width, color='b')
rects4 = ax.bar(ind+width*3, annual[3], width, color='k')
rects5 = ax.bar(ind+width*4, annual[4], width, color='c')
rects6 = ax.bar(ind+width*5, annual[5], width, color='k')
rects7 = ax.bar(ind+width*6, annual[6], width, color='r')
# add some text for labels, title and axes ticks
ax.set_ylabel('Percentage')
ax.set_title('ROE Annual')
ax.set_xticks(ind + width / 2)
ax.set_xticklabels(list(stock.loc['Fiscal Period'][astart:aend]))
#ax.legend((rects1[0], rects2[0]), ('workinprogress'))
def autolabel(rects, ax):
# Get y-axis height to calculate label position from.
(y_bottom, y_top) = ax.get_ylim()
y_height = y_top - y_bottom
for rect in rects:
height = rect.get_height()
# Fraction of axis height taken up by this rectangle
p_height = (height / y_height)
# If we can fit the label above the column, do that;
# otherwise, put it inside the column.
if p_height > 0.95: # arbitrary; 95% looked good to me.
label_position = height - (y_height * 0.05)
else:
label_position = height + (y_height * 0.01)
ax.text(rect.get_x() + rect.get_width()/2, label_position,
'%d' % int(height),
ha='center', va='bottom')
autolabel(rects1,ax)
autolabel(rects2,ax)
autolabel(rects3,ax)
autolabel(rects4,ax)
autolabel(rects5,ax)
autolabel(rects6,ax)
autolabel(rects7,ax)
plt.show()
我知道它现在还不是很漂亮,也没有懒惰地需要做更多的功能,但是似乎无法解决这个问题.感谢您的关注.
编辑:对于那些希望将来使用的人,这里的问题是S运算符,位于matplotlib documentation.我附上我的代码和新图表,因此可以轻松复制.仍然需要一些调整,但这是个人喜好.
astart = 21
aend = 31
annual = [Operating_Margin[astart:aend],Tax_Burden[astart:aend],Interest_Burden[astart:aend],Asset_Turnover[astart:aend],Leverage_Ratio[astart:aend],roe[astart:aend],Growth[astart:aend]]
N = len(annual[0])
ind = np.arange(N) # the x locations for the groups
width = .12 # the width of the bars
fig, ax = plt.subplots(figsize=(20,10),facecolor='#c8f2e5')
rects1 = ax.bar(ind, annual[0], width, color='#f29ca2')
rects2 = ax.bar(ind+width, annual[1], width, color='#61eaf2')
rects3 = ax.bar(ind+width*2, annual[2], width, color='#6da4d9')
rects4 = ax.bar(ind+width*3, annual[3], width, color='#f2bb12')
rects5 = ax.bar(ind+width*4, annual[4], width, color='c')
rects6 = ax.bar(ind+width*5, annual[5], width, color='#ce44f2')
rects7 = ax.bar(ind+width*6, annual[6], width, color='r')
ax.set_facecolor('#a7cff2')
# add some text for labels, title and axes ticks
ax.set_ylabel('Percentage',size=20)
ax.set_title('ROE Annual',size=30)
ax.set_xticks(ind + width / 2)
ax.set_xticklabels(list(stock.loc['Fiscal Period'][astart:aend]),size=14)
vals = ax.get_yticks()
ax.set_yticklabels(['{:3.2f}%'.format(x*100) for x in vals])
ax.legend((rects1[0], rects2[0], rects3[0], rects4[0], rects5[0], rects6[0], rects7[0]),('Operating Margin', 'Tax Burden','Interest Burden','Asset Turnover', 'Leverage Ratio','ROE','Growth'))
def autolabel(rects, ax):
# Get y-axis height to calculate label position from.
(y_bottom, y_top) = ax.get_ylim()
y_height = y_top - y_bottom
for rect in rects:
height = rect.get_height()
# Fraction of axis height taken up by this rectangle
p_height = (height / y_height)
# If we can fit the label above the column, do that;
# otherwise, put it inside the column.
if p_height > 0.95: # arbitrary; 95% looked good to me.
label_position = height - (y_height * 0.05)
else:
label_position = height + (y_height * 0.01)
ax.text(rect.get_x() + rect.get_width()/2, label_position,
'%.2f' % float(height),
ha='center', va='bottom',color='k',fontsize=12)
#
autolabel(rects1,ax)
autolabel(rects2,ax)
autolabel(rects3,ax)
autolabel(rects4,ax)
autolabel(rects5,ax)
autolabel(rects6,ax)
autolabel(rects7,ax)
plt.show()
解决方法:
我认为问题在于以下陈述.代替int,使用float
ax.text(rect.get_x() + rect.get_width()/2, label_position,
'%.2f' % float(height),
ha='center', va='bottom')
标签:matplotlib,python 来源: https://codeday.me/bug/20191111/2018343.html
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